Free Math Worksheets

- Where does a normal density curve attain its maximum? - How is the maximum height related to \(\sigma\)? - What does the axis of symmetry tell you about \(\mu\)?

1. The mean is the x-coordinate of the maximum and the axis of symmetry, so \(\mu=120\). 2. The maximum value of a normal density is \(f(\mu)=\frac{1}{\sigma\sqrt{2\pi}}\). Thus, \(0.08=\frac{1}{\sigma\sqrt{2\pi}}\). 3. Solving gives \(\sigma=\frac{1}{0.08\sqrt{2\pi}}\approx 4.99\).

\(\mu=120\) and \(\sigma\approx 4.99\)

- Compare the distances of \(45\) and \(55\) from the mean. - Relate the area left of one value to the area right of its reflection across the mean. - How are \(P(X\le\mu-a)\) and \(P(X\le\mu+a)\) related?

1. The values \(45=50-5\) and \(55=50+5\) are symmetric about the mean. 2. For a normal distribution, \(\Phi_{\mu,\sigma}(\mu-a)=1-\Phi_{\mu,\sigma}(\mu+a)\). 3. Therefore, \(\Phi_{50,\sigma}(55)\approx 1-0.3085=0.6915\).

\(\Phi_{50,\sigma}(55)\approx 0.6915\)

- What does the mean tell you about the center of a normal density curve? - Recall the empirical rule for values within two standard deviations of the mean. - At what distances from the mean does a normal density change concavity?

1. The mean \(\mu=250\) is the average active-ingredient content and the center of the symmetric distribution. 2. About \(95.4\%\) of a normal distribution lies within two standard deviations of the mean. The interval is \([250-2\cdot3, 250+2\cdot3]=[244, 256]\) milligrams. 3. A normal density has inflection points at \(\mu-\sigma\) and \(\mu+\sigma\). Therefore, the coordinates occur at \(x=247\) and \(x=253\).

1. \(\mu=250\,\text{mg}\) is the mean active-ingredient content. 2. Approximately \(95.4\%\) of the tablets contain between \(244\,\text{mg}\) and \(256\,\text{mg}\). 3. The inflection points occur at \(x=247\) and \(x=253\).

- How do probabilities at individual points differ for discrete and continuous random variables? - Use the symmetry of a normal distribution about its mean. - Consider how standard deviation affects the spread and height of the density curve.

1. A normal distribution is continuous. For any continuous random variable, the probability of one exact value is \(0\), because a single point has no area under the density curve. 2. Three hours is \(180\) minutes, which equals the mean. By symmetry, half of the distribution lies above the mean, so \(P(X>180)=0.5\). 3. Reducing the standard deviation concentrates the values more closely around the mean. The density curve becomes narrower and taller, while remaining centered at \(180\).

1. For a continuous distribution, the probability of any single exact value is \(0\). 2. \(50\%\) 3. The density curve becomes narrower and taller, with the same center.

- Compare the function with the general normal density formula. - Rewrite the exponent in the form \(-\frac{1}{2}(\cdots)^2\). - Check whether the coefficient agrees with the value of \(\sigma\). - The mean determines the location of the maximum, and the standard deviation determines the spread.

1. Compare the function with \(\varphi_{\mu,\sigma}(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\). 2. Rewrite the exponent: \(-0.02(x-12)^2=-\frac{1}{50}(x-12)^2=-\frac{1}{2}\left(\frac{x-12}{5}\right)^2\). 3. This identifies \(\mu=12\) and \(\sigma=5\). The coefficient also matches because \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{5\sqrt{2\pi}}\). Therefore, \(f\) is a normal density.

\(f\) is a normal density with \(\mu=12\) and \(\sigma=5\).

- The mean is the \(x\)-coordinate of the maximum and the line of symmetry. - The maximum height of a normal density is \(\frac{1}{\sigma\sqrt{2\pi}}\). - The inflection points are \(\sigma\) units to either side of the mean.

1. For graph \(f\), the maximum occurs at \(x=-1\), so \(\mu_1=-1\). The inflection points are at about \(-1.5\) and \(-0.5\), each \(0.5\) unit from the mean, so \(\sigma_1=0.5\). This also agrees with the maximum height \(\frac{1}{0.5\sqrt{2\pi}}\approx0.80\). 2. For graph \(g\), the maximum occurs at \(x=2\), so \(\mu_2=2\). The inflection points are at about \(1\) and \(3\), each \(1\) unit from the mean, so \(\sigma_2=1.0\). This agrees with the maximum height \(\frac{1}{\sqrt{2\pi}}\approx0.40\).

Graph \(f\): \(\mu_1=-1\), \(\sigma_1=0.5\) Graph \(g\): \(\mu_2=2\), \(\sigma_2=1.0\)

- Which parameter controls the horizontal location of a normal curve? - How does doubling the standard deviation change the height and width? - Compare curves with the same shape but different centers. - Compare curves with the same center but different spreads.

1. Graphs \(a\) and \(b\) are centered at \(x=0\), so they correspond to (I) and (II). Graph \(c\) is centered at \(x=3\), so it corresponds to (III). 2. Graph \(a\) is taller and narrower than graph \(b\), so graph \(a\) has the smaller standard deviation. Therefore, graph \(a\) corresponds to (I), and graph \(b\) corresponds to (II). 3. Graph \(c\) has the same shape as graph \(a\) but is shifted three units to the right, confirming that it corresponds to (III).

(I) Graph \(a\) (II) Graph \(b\) (III) Graph \(c\)

- Where does a normal density reach its maximum in terms of \(\mu\)? - How does the sign of the slope change on either side of the maximum? - Substitute \(x=\mu\) into the normal density formula.

1. a) A normal density reaches its maximum at \(x=\mu=4\). It is strictly increasing on \((-\infty, 4]\) and strictly decreasing on \([4, \infty)\). 2. b) The maximum value is \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{1.5\sqrt{2\pi}}\approx0.266\).

a) Strictly increasing on \((-\infty, 4]\); strictly decreasing on \([4, \infty)\) b) Approximately \(0.266\)

- Substitute \(\mu\) and \(\sigma\) into the formulas for the inflection-point inputs. - What does the exponent become when \(x=\mu+\sigma\)? - Symmetry gives both inflection points the same height.

1. The inflection-point inputs are \(x_1=\mu-\sigma=-3\) and \(x_2=\mu+\sigma=-1\). 2. By symmetry, both points have the same height. At either input, \(((x-\mu)/\sigma)^2=1\), so \(f(x)=\frac{1}{\sqrt{2\pi}}e^{-1/2}=\frac{1}{\sqrt{2\pi e}}\approx0.242\). 3. Therefore, \(W_1=\left(-3, \frac{1}{\sqrt{2\pi e}}\right)\) and \(W_2=\left(-1, \frac{1}{\sqrt{2\pi e}}\right)\).

\(W_1=\left(-3, \frac{1}{\sqrt{2\pi e}}\right)\), \(W_2=\left(-1, \frac{1}{\sqrt{2\pi e}}\right)\); the common \(y\)-coordinate is approximately \(0.242\).

- The mean is the \(x\)-coordinate of the maximum. - Locate the inflection points, where the curve changes concavity. - The distance from the mean to either inflection point equals \(\sigma\). - What happens to the exponent when \(x=\mu\)?

1. a) The maximum of \(f\) occurs at \(x=0\), so \(\mu_f=0\). Its inflection points are at \(-1\) and \(1\), so \(\sigma_f=1\). The maximum of \(g\) occurs at \(x=3\), so \(\mu_g=3\). Its inflection points are at \(1\) and \(5\), so \(\sigma_g=2\). 2. b) At the mean, the exponential factor equals \(1\). Thus, \(f(0)=\frac{1}{\sqrt{2\pi}}\approx0.40\), and \(g(3)=\frac{1}{2\sqrt{2\pi}}\approx0.20\). These values agree with the plotted maximum heights.

a) \(f\): \(\mu=0\), \(\sigma=1\); \(g\): \(\mu=3\), \(\sigma=2\) b) \(f(0)=\frac{1}{\sqrt{2\pi}}\approx0.40\); \(g(3)=\frac{1}{2\sqrt{2\pi}}\approx0.20\)

- Use the line of symmetry to locate the mean. - Identify the points where the curve is steepest; these are the inflection points. - Find the horizontal distance from each inflection point to the mean. - You may calculate the maximum height before applying the given relationship.

1. a) The maximum occurs at \(x=-1\), so \(\mu=-1\). The graph changes concavity at \(x=-2.5\) and \(x=0.5\). Each point is \(1.5\) units from the mean, so \(\sigma=1.5\). 2. b) The maximum height is \(h(-1)=\frac{1}{1.5\sqrt{2\pi}}\approx0.266\). Therefore, \(y_W\approx0.266e^{-0.5}\approx0.161\). This is consistent with the graph, where the height at \(x=-2.5\) and \(x=0.5\) is slightly above \(0.15\).

a) \(\mu=-1\), \(\sigma=1.5\) b) \(y_W\approx0.161\), which agrees with the graph.

- For a normal distribution, what cumulative probability occurs at the mean? - Use a difference of cumulative distribution values to find an interval probability. - A normal density is symmetric about its mean and reaches its maximum there.

1. a) For a normal distribution, \(F(\mu)=0.5\). The graph passes through \((4, 0.5)\), so \(\mu=4\). 2. b) From the graph, \(P(X\le4)=F(4)=0.5\). Also, \(P(2\le X\le6)=F(6)-F(2)\approx0.91-0.09=0.82\). 3. c) The density is a bell-shaped curve with its maximum at \(x=4\). It is symmetric about the line \(x=4\).

a) \(\mu=4\) b) \(P(X\le4)=0.5\); \(P(2\le X\le6)\approx0.82\) c) The density has its maximum at \(x=4\) and is symmetric about \(x=4\).

- Rewrite the exponent as a constant times a perfect square. - Compare the result with the general normal density formula. - Pay attention to how the coefficient of the squared expression determines \(\sigma\).

1. Complete the square in the exponent: \(-0.5x^2+4x-8=-0.5(x^2-8x+16)=-0.5(x-4)^2\). 2. Therefore, \(f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-4)^2}{2}}\). Comparing this with \(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\) gives \(\mu=4\) and \(\sigma=1\). The leading factor also matches \(\frac{1}{1\sqrt{2\pi}}\), confirming that \(f\) is a normal density.

\(f(x)=\frac{1}{\sqrt{2\pi}}e^{-0.5(x-4)^2}\), so \(\mu=4\) and \(\sigma=1\).

- Locate the line of symmetry of the density curve. - The inflection points are one standard deviation from the mean. - Probabilities for continuous random variables are represented by areas.

1. a) The density curve reaches its maximum at \(x=4\), so \(\mu=4\). 2. b) The inflection points occur at \(x=2.5\) and \(x=5.5\). Their distance from the mean is \(1.5\), so \(\sigma=1.5\). 3. c) A normal random variable is continuous. The probability of any one exact value is \(0\), because a single point has zero width and therefore zero area under the density curve. Thus, \(P(X=4)=0\).

a) \(\mu=4\) b) \(\sigma=1.5\) c) \(P(X=4)=0\)

- The center of a normal density is its mean. - Compare the widths and maximum heights of the two curves. - Interpret each probability as an area under its density curve.

1. a) The mean is the \(x\)-coordinate of the maximum. The maxima occur at \(2\) for \(f_1\) and \(4\) for \(f_2\), so \(\mu_2>\mu_1\). 2. b) The graph of \(f_2\) is wider and lower than the graph of \(f_1\), so \(\sigma_2>\sigma_1\). Thus, \(X_2\) has greater spread. 3. c) Probability is area under the density curve. On \([1.5, 2.5]\), \(f_1\) is much higher than \(f_2\), so \(P(1.5\le X_1\le2.5)>P(1.5\le X_2\le2.5)\).

a) \(\mu_2>\mu_1\), since \(4>2\) b) \(\sigma_2>\sigma_1\); \(X_2\) has greater spread. c) The probability is greater for \(X_1\).

- Compare the \(x\)-coordinates of the maxima. - A flatter curve represents greater spread. - Use the formula for the maximum height or the fact that total area remains \(1\).

1. a) The dashed curve reaches its maximum at \(x=30\), so \(\mu_Y=30\). Its maximum is about half as high as that of \(X\), and its curve is about twice as wide. Since the maximum height is inversely proportional to \(\sigma\), \(\sigma_Y=2\cdot5=10\). 2. b) Increasing the mean shifts the entire graph to the right without changing its shape. 3. c) The maximum height is \(\frac{1}{\sigma\sqrt{2\pi}}\). Therefore, a larger standard deviation produces a lower and wider density curve.

a) \(\mu_Y=30\), \(\sigma_Y=10\) b) Changing the mean shifts the graph horizontally without changing its shape. c) As the standard deviation increases, the maximum height decreases.

- The cumulative distribution function value \(F(k)\) equals \(P(X\le k)\). - Find an interval probability by subtracting cumulative probabilities at the endpoints. - Use the complement for a “greater than” probability.

1. a) \(P(X\le4)=F(4)=0.5\). 2. b) \(P(X\le5)=F(5)\approx0.84\). 3. c) \(P(3\le X\le5)=F(5)-F(3)\approx0.84-0.16=0.68\). 4. d) \(P(X>6)=1-F(6)\approx1-0.98=0.02\).

a) \(P(X\le4)=0.5\) b) \(P(X\le5)\approx0.84\) c) \(P(3\le X\le5)\approx0.68\) d) \(P(X>6)\approx0.02\)

- The maximum of a normal density occurs at the mean. - Compare the widths and heights of the curves. - Does a horizontal translation change spread?

1. a) The mean is the \(x\)-coordinate of the maximum. The maxima occur at \(x=3\) and \(x=7\), so \(\mu_1=3\) and \(\mu_2=7\). 2. b) The graph of \(f_2\) is the graph of \(f_1\) shifted \(4\) units to the right. The curves have the same height and width, so their standard deviations are equal: \(\sigma_1=\sigma_2\).

a) \(\mu_1=3\), \(\mu_2=7\) b) Shift \(f_1\) four units to the right. Therefore, \(\sigma_1=\sigma_2\).

- What does standard deviation indicate about the spread of a normal curve? - How are width and height related when the total area must remain constant? - Recall the defining total-area condition for a density function.

1. a) The graph of \(h_2\) is wider and flatter than the graph of \(h_1\). Greater spread around the mean means a larger standard deviation, so \(\sigma_2>\sigma_1\). 2. b) The total area under every density curve must equal \(1\). When a normal curve spreads over a wider range, its height must decrease so that the area remains \(1\). Therefore, the curve with the larger standard deviation has the lower maximum.

a) \(h_2\) has the larger standard deviation because its curve is wider. b) Both curves must have total area \(1\), so a wider curve must be lower to preserve that area.

- Compare the expression with the general normal density formula. - Identify which parameter controls the center and which controls the width. - A horizontal stretch of a density requires a corresponding vertical compression to preserve total area. - Track how the maximum point of the standard normal curve changes under each transformation.

1. Comparing the function with \(\phi_{\mu,\sigma}(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12((x-\mu)/\sigma)^2}\) gives \(\mu=-4\) and \(\sigma=2.5\). Starting with the standard normal graph, stretch horizontally by a factor of \(2.5\), compress vertically by a factor of \(\frac{1}{2.5}=0.4\), and shift \(4\) units left. 2. The maximum occurs at \(x=\mu=-4\). Its height is \(\frac{1}{2.5\sqrt{2\pi}}\), so the maximum point is \(\left(-4, \frac{1}{2.5\sqrt{2\pi}}\right)\).

1. Horizontal stretch by \(2.5\), vertical compression by \(0.4\), and shift \(4\) units left 2. \(\left(-4, \frac{1}{2.5\sqrt{2\pi}}\right)\)

- How does a horizontal stretch affect area under a graph? - Relate the horizontal stretch factor to the standard deviation. - Which coordinate of the maximum is determined by the horizontal shift? - Use the maximum height to identify the spread.

1. The horizontal shift places the center and maximum at \(x=5\), so \(\mu=5\). 2. The maximum height of a normal density is \(\frac{1}{\sigma\sqrt{2\pi}}\). Comparing this with \(\frac{1}{4\sqrt{2\pi}}\) gives \(\sigma=4\). 3. The horizontal stretch factor is \(k=\sigma=4\). To preserve area, the vertical scale factor must be its reciprocal, \(\frac14=0.25\).

\(\mu=5\), \(\sigma=4\), and the vertical scale factor is \(\frac14=0.25\).

- Compare the expression with the general formula for a normal density. - Which value identifies the center of the normal curve? - Relate the denominator in the exponent to \(2\sigma^2\). - Standardize the labeled weight before using the standard normal distribution.

1. a) Compare the density with \(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\). The center is \(\mu=1.02\). Also, \(2\sigma^2=0.0008\), so \(\sigma^2=0.0004\) and \(\sigma=0.02\). The coefficient agrees because \(\sigma\sqrt{2\pi}=\sqrt{0.0008\pi}\). 2. b) Standardize \(1.00\): \(z=\frac{1.00-1.02}{0.02}=-1\). Therefore, \(P(X<1.00)=\Phi(-1)\approx 0.1587\).

a) \(\mu=1.02\,\text{kg}\) and \(\sigma=0.02\,\text{kg}\) b) \(P(X<1.00)\approx 0.1587\), or about \(15.87\%\)

- Read each mean from the x-coordinate of the maximum point. - Use the maximum-height formula to find each standard deviation. - How does a larger standard deviation affect the width and height of a normal curve? - The total area under each density curve remains \(1\).

1. a) For \(X\), the maximum occurs at the mean, so \(\mu_X=10\). Since \(0.2=\frac{1}{\sigma_X\sqrt{2\pi}}\), \(\sigma_X=\frac{1}{0.2\sqrt{2\pi}}\approx 1.99\). 2. For \(Y\), \(\mu_Y=20\). Since \(0.05=\frac{1}{\sigma_Y\sqrt{2\pi}}\), \(\sigma_Y=\frac{1}{0.05\sqrt{2\pi}}\approx 7.98\). 3. b) A larger standard deviation produces a wider, flatter normal curve. Since \(\sigma_Y>\sigma_X\), the density curve for \(Y\) is flatter.

a) \(X\): \(\mu_X=10\), \(\sigma_X\approx 1.99\); \(Y\): \(\mu_Y=20\), \(\sigma_Y\approx 7.98\) b) The curve for \(Y\) is flatter because \(\sigma_Y>\sigma_X\).

- Express an interval probability as a difference of cumulative probabilities. - Replace the term containing \(\mu-a\) using the given identity. - Pay attention to signs when removing parentheses.

1. Write the interval probability as \(P(\mu-a\le X\le\mu+a)=\Phi_{\mu,\sigma}(\mu+a)-\Phi_{\mu,\sigma}(\mu-a)\). 2. Substitute \(\Phi_{\mu,\sigma}(\mu-a)=1-\Phi_{\mu,\sigma}(\mu+a)\). 3. Then \(P=\Phi_{\mu,\sigma}(\mu+a)-[1-\Phi_{\mu,\sigma}(\mu+a)]=2\Phi_{\mu,\sigma}(\mu+a)-1\).

\(P(\mu-a\le X\le\mu+a)=2\Phi_{\mu,\sigma}(\mu+a)-1\)

- How do \(\mu\) and \(\sigma\) affect the location and spread of a normal curve? - If a normal curve becomes narrower while its total area stays \(1\), what happens to its maximum height? - Where are the inflection points located relative to \(\mu\)? - Use the formula for the density at \(x=\mu\).

1. a) Since \(\mu=4\), the graph is shifted \(4\) units to the right. Since \(\sigma=0.5<1\), the graph is narrower and taller than the standard normal curve. 2. b) The maximum occurs at \(x=\mu=4\). Its value is \(f(4)=\frac{1}{0.5\sqrt{2\pi}}=\frac{2}{\sqrt{2\pi}}\approx 0.7979\). The standard normal maximum is \(\frac{1}{\sqrt{2\pi}}\approx 0.3989\), so the maximum of \(f\) is exactly twice as high. 3. c) The inflection points of a normal density occur at \(x=\mu\pm\sigma\). Thus, the x-coordinates are \(3.5\) and \(4.5\).

a) The graph is shifted \(4\) units right and is narrower and taller than the standard normal curve. b) The maximum point is \(\left(4, \frac{2}{\sqrt{2\pi}}\right)\approx (4, 0.7979)\). Its y-coordinate is twice the standard normal maximum. c) The inflection points have x-coordinates \(3.5\) and \(4.5\).

- What function-value equation proves symmetry about the y-axis? - Changes inside a function affect horizontal transformations. - How does translating an entire graph affect its symmetry axis?

1. a) \(\varphi_{0,1}(-x)=\frac{1}{\sqrt{2\pi}}e^{-(-x)^2/2}=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}=\varphi_{0,1}(x)\). Therefore, the graph is symmetric about the y-axis. 2. b) First, stretch the graph horizontally by a factor of \(\sigma\). Next, shift it horizontally by \(\mu\) units. Finally, scale it vertically by a factor of \(\frac{1}{\sigma}\) so the total area remains \(1\). 3. c) The original symmetry axis is \(x=0\). Horizontal and vertical scaling leave this vertical axis in place, while the horizontal shift by \(\mu\) moves it to \(x=\mu\).

a) \(\varphi_{0,1}(-x)=\varphi_{0,1}(x)\) b) Apply a horizontal scale factor of \(\sigma\), shift horizontally by \(\mu\), and apply a vertical scale factor of \(\frac{1}{\sigma}\). c) The shift moves the symmetry axis from \(x=0\) to \(x=\mu\).

- Simplify the fraction inside the squared expression. - Compare the result with the general normal density formula. - Rewrite the coefficient to check whether it equals \(\frac{1}{\sigma\sqrt{2\pi}}\). - The total area is \(1\) only when the coefficient and standard deviation agree.

1. Simplify the expression in the exponent: \(\frac{2x-10}{8}=\frac{2(x-5)}{8}=\frac{x-5}{4}\). 2. Then \(g(x)=\frac{1}{\sqrt{32\pi}}e^{-\frac{1}{2}\left(\frac{x-5}{4}\right)^2}\), so comparison with the normal density formula gives \(\mu=5\) and \(\sigma=4\). 3. The coefficient is correct because \(\frac{1}{4\sqrt{2\pi}}=\frac{1}{\sqrt{32\pi}}\). Therefore, \(g\) is a normal density.

\(g\) is a normal density with \(\mu=5\) and \(\sigma=4\).

- Where does a normal density curve reach its maximum? - How is the height of that maximum related to \(\sigma\)? - Compare each endpoint of the interval with the mean in units of standard deviation.

1. A normal density reaches its maximum at \(x=\mu\), so \(\mu=4\). The maximum value is \(\frac{1}{\sigma\sqrt{2\pi}}\). Thus, \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{\sqrt{18\pi}}\). Squaring gives \(2\pi\sigma^2=18\pi\), so \(\sigma^2=9\) and \(\sigma=3\). 2. The interval \([1, 7]\) is \([\mu-\sigma, \mu+\sigma]\). Therefore, \(P(1\le X\le7)=P(-1\le Z\le1)\approx0.6827\).

1. \(\mu=4\), \(\sigma=3\) 2. \(P(1\le X\le7)\approx0.6827\), or about \(68.27\%\)

- Sketch a normal curve and mark the mean. - Reflect the right-tail probability across the mean to an equivalent left-tail probability. - Use the fact that a normal cumulative distribution function is strictly increasing.

1. Since \(150=120+30\), symmetry about \(\mu=120\) gives \(P(X\ge 150)=P(X\le 120-30)=P(X\le 90)\). 2. The inequality becomes \(P(X\le 120+k)\le P(X\le 90)\). 3. A normal cumulative distribution function is strictly increasing, so \(120+k\le 90\). Therefore, \(k\le -30\).

\(k\le -30\)

- The mean is located at the maximum and on the line of symmetry. - The inflection points are one standard deviation from the mean. - Substitute \(x=\mu\) into the general normal density formula. - The total area must stay equal to \(1\) as the curve becomes wider.

1. a) The mean is the \(x\)-coordinate of the maximum. Thus, \(\mu_1=-2\), \(\mu_2=3\), and \(\mu_3=0\). The inflection points lie \(\sigma\) units from the mean. From the graph, the corresponding distances are \(1\), \(2\), and \(4\), so \(\sigma_1=1\), \(\sigma_2=2\), and \(\sigma_3=4\). 2. b) For \(f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12((x-\mu)/\sigma)^2}\), the exponential factor equals \(1\) at \(x=\mu\). Therefore, the maximum is \(f(\mu)=\frac{1}{\sigma\sqrt{2\pi}}\), which is inversely proportional to \(\sigma\). A larger standard deviation produces a lower, wider curve.

a) \(X_1\): \(\mu_1=-2\), \(\sigma_1=1\) \(X_2\): \(\mu_2=3\), \(\sigma_2=2\) \(X_3\): \(\mu_3=0\), \(\sigma_3=4\) b) The maximum is \(f(\mu)=\frac{1}{\sigma\sqrt{2\pi}}\), so it decreases as \(\sigma\) increases.

- What symmetry and shape properties must every normal density have? - How are a normal curve's inflection points related to \(\sigma\)? - How is the maximum height related to \(\sigma\)? - The total area under a density curve must equal \(1\).

1. a) The graph \(g\) is symmetric about \(x=2\) and has inflection points at \(x=1\) and \(x=3\), which would imply \(\sigma=1\). A normal density with \(\sigma=1\) must have maximum height \(\frac{1}{\sqrt{2\pi}}\approx 0.40\), but \(g\) has maximum height \(0.6\), so it is not normalized. 2. The graph \(h\) is not symmetric about its maximum, while every normal density is symmetric about its mean. 3. b) The graph \(f\) has its maximum and axis of symmetry at \(x=2\), so \(\mu=2\). Its inflection points are at \(x=1\) and \(x=3\), one unit from the mean, so \(\sigma=1\). Its maximum height also agrees with \(\frac{1}{\sqrt{2\pi}}\approx 0.40\).

a) The graph \(g\) has the wrong maximum height for the standard deviation indicated by its inflection points. The graph \(h\) is asymmetric, so it cannot be a normal density. b) For the graph \(f\), \(\mu=2\) and \(\sigma=1\).

- What is the cumulative probability at the mean of a normal distribution? - What cumulative probabilities occur near \(\mu-\sigma\) and \(\mu+\sigma\)? - Express an interval probability as a difference of cumulative distribution values.

1. a) The mean is where \(F(\mu)=0.5\). The graph gives \(\mu=4\). For a normal distribution, \(F(\mu+\sigma)\approx 0.8413\). The graph reaches about \(0.84\) at \(x=6\), so \(\sigma=6-4=2\). 2. b) Read \(F(6)\approx 0.84\) and \(F(2)\approx 0.16\). Then \(P(2\le X\le 6)=F(6)-F(2)\approx 0.84-0.16=0.68\).

a) \(\mu=4\) and \(\sigma=2\) b) \(P(2\le X\le 6)\approx 0.68\), or about \(68\%\)

- The mean is at the maximum and on the line of symmetry. - The inflection points are one standard deviation from the mean. - The total area is \(1\), and symmetry divides it equally at the mean. - Express \([3, 7]\) in terms of \(\mu_2\) and \(\sigma_2\), then use the empirical rule.

1. a) The mean is the \(x\)-coordinate of the maximum, and the inflection points are one standard deviation from the mean. For \(f\), the maximum is at \(2\) and the inflection points are at \(1\) and \(3\), so \(\mu_1=2\) and \(\sigma_1=1\). For \(g\), the maximum is at \(5\) and the inflection points are at \(3\) and \(7\), so \(\mu_2=5\) and \(\sigma_2=2\). 2. b) By symmetry, half of each normal distribution lies on either side of its mean. Therefore, \(P(X_1\le2)=0.5\) and \(P(X_2\ge5)=0.5\). 3. c) The interval \([3, 7]\) is \([\mu_2-\sigma_2, \mu_2+\sigma_2]\). By the empirical rule, \(P(3\le X_2\le7)\approx0.683\).

a) \(\mu_1=2\), \(\sigma_1=1\); \(\mu_2=5\), \(\sigma_2=2\) b) (1) \(P(X_1\le2)=0.5\) (2) \(P(X_2\ge5)=0.5\) c) \(P(3\le X_2\le7)\approx0.683\), or about \(68.3\%\)

- The maximum of a normal density identifies the mean. - The inflection points are one standard deviation from the mean. - Express the endpoints using \(\mu\) and \(\sigma\), then apply the empirical rule. - A larger standard deviation spreads the same total area over a wider interval.

1. a) The maximum occurs at \(x=10\), so \(\mu=10\). The inflection points are at \(x=8\) and \(x=12\), each \(2\) units from the mean, so \(\sigma=2\). 2. b) The interval \([6, 14]\) is \([\mu-2\sigma, \mu+2\sigma]\). By the empirical rule, \(P(6\le X\le14)\approx0.9545\). 3. c) Doubling the standard deviation to \(4\) makes the curve wider and lower while leaving its center at \(10\). The new maximum is \(\left(10, \frac{1}{4\sqrt{2\pi}}\right)\approx(10, 0.100)\). The new inflection points occur at \(x=10-4=6\) and \(x=10+4=14\).

a) \(\mu=10\), \(\sigma=2\) b) \(P(6\le X\le14)\approx0.9545\), or about \(95.45\%\); this is the interval within two standard deviations of the mean. c) The curve becomes wider and lower. Its new maximum is approximately \((10, 0.100)\), and its inflection points occur at \(x=6\) and \(x=14\).

- At what \(x\)-value does a normal cumulative distribution function equal \(0.5\)? - How does greater spread affect the steepness of a cumulative distribution function? - Use a secant line through points close to the inflection point.

1. a) Both cumulative distribution functions have value \(0.5\) at \(x=2\). Therefore, \(\mu_1=\mu_2=2\). 2. b) The graph of \(F_1\) is steeper near the mean, so \(X_1\) has the smaller standard deviation. Therefore, \(X_2\) has the larger standard deviation. 3. c) Using points near the inflection point, \(F_2(1.5)\approx0.37\) and \(F_2(2.5)\approx0.63\). The secant slope is \(\frac{0.63-0.37}{2.5-1.5}=0.26\), which estimates the maximum slope. Thus, \(\sigma_2\approx\frac{0.4}{0.26}\approx1.54\), or about \(1.5\).

a) \(\mu_1=\mu_2=2\) b) \(X_2\) has the larger standard deviation because \(F_2\) rises more gradually. c) The maximum slope is about \(0.26\), giving \(\sigma_2\approx1.5\).

- Rewrite the exponent in the form \(-c(x-x_0)^2\). - Relate the coefficient \(c\) to \(\frac{1}{2\sigma^2}\). - Check that the leading factor agrees with the value of \(\sigma\).

1. Complete the square: \(-4.5x^2+9x-4.5=-4.5(x^2-2x+1)=-4.5(x-1)^2\). 2. Compare \(-4.5(x-1)^2\) with \(-\frac{(x-\mu)^2}{2\sigma^2}\). This gives \(\mu=1\) and \(\frac{1}{2\sigma^2}=4.5\), so \(\sigma^2=\frac19\) and \(\sigma=\frac13\). 3. The required leading factor is \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{3}{\sqrt{2\pi}}\), which matches the given function. Thus, \(g\) is a normal density, and its maximum occurs at \(x=\mu=1\).

\(g\) is the density of a normal distribution with \(\mu=1\) and \(\sigma=\frac13\). Its maximum occurs at \(x=1\).

- What value does a normal cumulative distribution function have at the mean? - Recall the cumulative probability at one standard deviation above the mean. - Use the normal density formula with the values of \(\mu\) and \(\sigma\). - Where does a normal density function reach its maximum?

1. a) The mean is the \(x\)-value for which \(F(x)=0.5\). From the graph, \(\mu=4\). 2. b) For a normal distribution, \(F(\mu+\sigma)\approx0.8413\). The graph shows \(F(5.5)\approx0.84\), so \(4+\sigma=5.5\) and \(\sigma=1.5\). 3. c) The density reaches its maximum at the mean. Its value is \(f(4)=\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{1.5\sqrt{2\pi}}\approx0.266\).

a) \(\mu=4\) b) \(\sigma=1.5\) c) \(f(4)\approx0.266\)

- A cumulative probability is represented by area under the density curve to the left of a value. - How much of a symmetric normal distribution lies to the left of its mean? - Add the estimated area from \(500\) to \(510\) to the area left of \(500\). - Consider where a normal cumulative distribution function begins, ends, and crosses \(0.5\). - Think about what happens when values are more tightly concentrated around the mean.

1. a) A normal distribution is symmetric about its mean, so half of the total area lies to the left of \(500\). Thus, \(P(X\le500)=0.5\). 2. b) The area from \(500\) to \(510\) is about \(6.8\) grid squares, or \(6.8\cdot0.05=0.34\). Therefore, \(F(510)\approx0.5+0.34=0.84\). 3. c) The cumulative distribution function is S-shaped, approaches \(0\) for small \(x\), and approaches \(1\) for large \(x\). It passes through \((500, 0.5)\). 4. d) A smaller standard deviation makes the density curve narrower and taller while keeping its center at \(500\).

a) \(P(X\le500)=0.5\) b) \(F(510)\approx0.84\) c) An S-shaped curve through \((500, 0.5)\) d) The density curve becomes narrower and taller, with the same center.

- The mean occurs where a normal cumulative distribution function equals \(0.5\). - A steeper cumulative distribution function indicates less spread. - Relate the input where the cumulative probability is \(0.841\) to \(\mu+\sigma\).

1. a) A normal cumulative distribution function equals \(0.5\) at the mean. The graph shows \(\mu_1\approx10.0\,\text{mm}\) and \(\mu_2\approx10.5\,\text{mm}\), so Machine 2 has the larger mean diameter. 2. b) A smaller standard deviation gives a steeper cumulative distribution function near the mean. Since \(F_2\) is steeper, Machine 2 is more precise. 3. c) For Machine 1, \(F_1(x)\approx0.841\) at \(x\approx10.4\). Since this input equals \(\mu_1+\sigma_1\), \(\sigma_1\approx10.4-10.0=0.4\,\text{mm}\).

a) Machine 2 b) Machine 2 c) \(\sigma_1\approx0.4\,\text{mm}\)

- Complete the square in the exponent. - Compare the result with the general normal density formula. - Check that the value of \(\sigma\) also matches the coefficient in front of the exponential.

1. Complete the square in the exponent: \(-\frac{1}{8}x^2+\frac{1}{2}x-\frac{1}{2}=-\frac{1}{8}(x^2-4x+4)=-\frac{1}{8}(x-2)^2\). 2. Compare with the normal exponent \(-\frac{(x-\mu)^2}{2\sigma^2}\). Since \(-\frac{1}{8}(x-2)^2=-\frac{(x-2)^2}{2\cdot 4}\), \(\mu=2\) and \(\sigma^2=4\). 3. Therefore, \(\sigma=2\). The coefficient checks because \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{2\sqrt{2\pi}}=\frac{1}{\sqrt{8\pi}}\).

\(\mu=2\) and \(\sigma=2\)

- Complete the square so the exponent matches the standard form of a normal density. - How is the coefficient related to \(\sigma\)? - Recall the relationship between \(\mu\), \(\sigma\), and the inflection points of a normal curve. - The total area under a probability density must equal \(1\).

1. Complete the square in the exponent: \(-0.125x^2+0.5x-0.5=-0.125(x^2-4x+4)=-0.125(x-2)^2\). 2. Compare with \(-\frac{(x-\mu)^2}{2\sigma^2}\). This gives \(\mu=2\) and \(\frac{1}{2\sigma^2}=0.125=\frac{1}{8}\), so \(\sigma^2=4\) and \(\sigma=2\). 3. A normal density has coefficient \(c=\frac{1}{\sigma\sqrt{2\pi}}\), so \(c=\frac{1}{2\sqrt{2\pi}}\). 4. A normal density has inflection points at \(x=\mu\pm\sigma\). Therefore, the x-coordinates are \(2-2=0\) and \(2+2=4\).

\(c=\frac{1}{2\sqrt{2\pi}}\); the inflection points have x-coordinates \(0\) and \(4\).

- To prove symmetry about \(x=a\), compare function values at \(a-h\) and \(a+h\). - Where does a normal density attain its maximum? - Examine how \(\sigma\) appears in the coefficient of the density function.

1. a) For any real \(h\), substitute \(\mu-h\) and \(\mu+h\). The exponents contain \(\left(\frac{-h}{\sigma}\right)^2\) and \(\left(\frac{h}{\sigma}\right)^2\), which are equal. Therefore, \(\varphi_{\mu,\sigma}(\mu-h)=\varphi_{\mu,\sigma}(\mu+h)\), proving symmetry about \(x=\mu\). 2. b) The maximum occurs at \(x=0\) and equals \(\frac{1}{\sigma\sqrt{2\pi}}\). Solving \(\frac{1}{\sigma\sqrt{2\pi}}\approx 0.133\) gives \(\sigma\approx \frac{1}{0.133\sqrt{2\pi}}\approx 3.00\). 3. c) The maximum value is inversely proportional to \(\sigma\). Replacing \(\sigma\) with \(4\sigma\) changes the maximum to \(\frac{1}{4\sigma\sqrt{2\pi}}\), one-fourth of its original value.

a) \(\varphi_{\mu,\sigma}(\mu-h)=\varphi_{\mu,\sigma}(\mu+h)\), so the graph is symmetric about \(x=\mu\). b) \(\sigma\approx 3.00\) c) The maximum value is divided by \(4\).