Free math worksheets

- Compare the value’s deviation from the mean with the standard deviation. - Use \(z=\frac{x-\mu}{\sigma}\), then interpret the sign and magnitude in standard-deviation units.

1. \(z=\frac{86-70}{8}=2\). 2. The value \(86\) is \(2\) standard deviations above the population mean.

\(z=2\); the value is \(2\) standard deviations above the mean.

- Express each endpoint as a number of standard deviations from the mean. - Match the interval to the empirical-rule band centered at the mean.

1. The interval \([44,56]\) is within one standard deviation of the mean. 2. The empirical rule assigns approximately \(68\%\) to this interval.

Approximately \(68\%\).

- Relate standard deviation to horizontal spread while preserving total area. - Imagine narrowing a curve while keeping its area equal to \(1\); infer what must happen to its height.

1. A smaller standard deviation means values are more concentrated near the mean. 2. The corresponding normal curve is taller and narrower. 3. Curve A has the smaller standard deviation.

Curve A is taller and more concentrated around the mean.

- Work backward from the standardized position to the original scale. - Rearrange the z-score relationship as \(x=\mu+z\sigma\) and preserve the negative sign.

1. Use \(x=\mu+z\sigma\). 2. \(x=120+(-1.4)\cdot15=99\).

The value is \(99\).

- Put both performances on the same standardized scale. - The larger z-score represents the stronger performance relative to its own test distribution.

1. Student A: \(z_A=\frac{78-70}{4}=2\). 2. Student B: \(z_B=\frac{640-600}{50}=0.8\). 3. Since \(2>0.8\), Student A had the higher relative position.

Student A, with \(z=2\) compared with Student B’s \(z=0.8\).

- Express each endpoint as a number of standard deviations from the mean. - Choose the corresponding central empirical-rule percentage, then use its complement.

1. The endpoints are two standard deviations from the mean. 2. Approximately \(95\%\) lies within two standard deviations. 3. The percentage outside is \(100\%-95\%=5\%\).

Approximately \(5\%\).

- Use nested central regions and symmetry. - Subtract the one-standard-deviation central area from the two-standard-deviation central area, then use symmetry.

1. Approximately \(95\%\) lies within two standard deviations and \(68\%\) within one. 2. The area between one and two standard deviations on both sides is \(95\%-68\%=27\%\). 3. By symmetry, the upper-side portion is \(\frac{27\%}{2}=13.5\%\).

Approximately \(13.5\%\).

- Standardize the boundary and identify which side of it is requested. - After standardizing, use normal CDF technology for the cumulative area to the left.

1. Standardize: \(z=\frac{85-72}{10}=1.30\). 2. The standard normal area to the left of \(1.30\) is \(0.9032\).

\(P(X<85)\approx 0.9032\).

- Standardize the cutoff and use the tail on the correct side. - A greater-than probability is an upper tail, so take the complement of the left-tail cumulative area.

1. \(z=\frac{560-500}{40}=1.50\). 2. The upper-tail probability is \(1-\Phi(1.50)=0.0668\).

The probability is approximately \(0.0668\).

- Convert both endpoints to the standard scale and find the area between them. - Find the cumulative area at each standardized endpoint and subtract the lower value from the upper value.

1. The standardized bounds are \(-1.20\) and \(1.20\). 2. Subtract the left-tail areas: \(\Phi(1.20)-\Phi(-1.20)=0.7699\).

\(P(24<X<36)\approx 0.7699\).

- Compare distances from the means in standard-deviation units. - Compare absolute z-scores because unusualness depends on distance from the mean, not direction.

1. In A, \(z=\frac{15}{5}=3\). 2. In B, \(z=\frac{15}{10}=1.5\). 3. The larger absolute z-score occurs in A, so \(85\) is more unusual there.

Population A, because \(z=3\) there compared with \(z=1.5\) in Population B.

- Identify how many standard deviations the interval extends, then use the complementary percentage. - Use the empirical-rule percentage outside three standard deviations, then apply it to the sample size and round to a whole observation.

1. The interval endpoints are \(3\) standard deviations from the mean. 2. Approximately \(99.7\%\) lie inside, so \(0.3\%\) lie outside. 3. The expected outside count is \(0.003\cdot1200=3.6\), about \(4\).

About \(4\) observations.

- Subtract the cumulative area at the lower boundary from that at the upper boundary. - Use normal CDF technology for both cumulative areas, then subtract.

1. The left-tail areas are \(\Phi(1)\approx 0.84134\) and \(\Phi(-2)\approx 0.02275\). 2. The area between is \(0.84134-0.02275=0.81859\approx 0.8186\).

\(P(-2<Z<1)\approx 0.8186\).

- Use the standardized-score equation and solve for the scale parameter. - Substitute the known mean, value, and z-score into the formula, then solve for the positive standard deviation.

1. Use \(-1.5=\frac{30-42}{\sigma}\). 2. Then \(-1.5=\frac{-12}{\sigma}\), so \(\sigma=8\).

\(\sigma=8\).

- Distinguish a raw difference from a standardized difference. - Ask what quantity sets the scale for comparing deviations. - Check what happens to the units when one measured difference is divided by another.

1. The student found the raw deviation but did not express it in standard-deviation units. 2. The correct calculation is \(z=\frac{82-70}{6}=2\). 3. Both the numerator and denominator are measured in score points, so their units cancel.

The student forgot to divide by the standard deviation. The correct standard score is \(z=2\), and it is unitless because the score units cancel in the quotient.

- Interpret each percentile as cumulative area to the left. - Find the area between two cumulative positions by subtraction. - Decide whether any feature specific to a normal curve was used.

1. The area to the left of \(a\) is \(0.18\), and the area to the left of \(b\) is \(0.73\). 2. The proportion between them is \(0.73-0.18=0.55\). 3. The calculation uses only percentile definitions, so normality is not required.

The proportion is \(0.55\), or \(55\%\). The result does not require a normal distribution.

- Identify the shape condition required by the stated percentage rule. - Separate what the mean and standard deviation locate from what they reveal about area. - Decide whether the distribution's described shape meets the rule's assumptions.

1. The interval \([40,60]\) is one standard deviation on either side of the mean. 2. The \(68\%\) empirical-rule estimate requires an approximately bell-shaped distribution. 3. A strongly bimodal distribution does not satisfy that shape condition, so the percentage cannot be inferred from the mean and standard deviation alone.

The claim is not justified. Although \([40,60]\) is within one standard deviation of the mean, the empirical rule does not apply to a strongly bimodal distribution.

- Think about how cumulative area changes as a cutoff moves to the right. - Compare the ordering of the two raw values with the ordering of their claimed percentiles. - No normal-table calculation is needed to test the logical consistency.

1. Percentile rank is the cumulative proportion at or below a value. 2. Within one continuous distribution, a larger raw value must have a percentile rank at least as large as that of a smaller raw value. 3. Because \(50>42\) but \(60\%<70\%\), the reported ordering is impossible.

The claims are inconsistent. In the same distribution, the larger value \(50\) cannot have a lower percentile rank than the smaller value \(42\).

- Compare the two percentile percentages through normal-curve symmetry. - Find the given value's raw distance from the mean. - Reflect that distance to the other side of the center.

1. The \(12\)th and \(88\)th percentiles are symmetric because their cumulative probabilities add to \(1\). 2. The value \(34\) is \(50-34=16\) units below the mean. 3. The symmetric value is \(50+16=66\).

The \(88\)th-percentile value is \(66\).

- Translate the standardized position into the relevant one-sided tail area. - Compare the resulting percentage with the stated decision threshold. - Keep the direction of the tail consistent with the negative standard score.

1. The lower-tail probability is \(P(Z<-2.4)\approx0.0082\). 2. As a percentage, this is approximately \(0.82\%\). 3. Because \(0.82\%<1\%\), the measurement meets the flagging rule.

Yes. Its lower-tail probability is approximately \(0.0082\), or \(0.82\%\), which is below the \(1\%\) threshold.

- Undo the linear transformation used to create the reported score. - Interpret the resulting value on the ordinary standard normal scale. - Convert the cumulative area to a percentile percentage.

1. Solve \(68=50+10z\), giving \(z=1.8\). 2. The cumulative standard normal area at \(z=1.8\) is approximately \(0.9641\).

The ordinary standard score is \(z=1.8\), corresponding to a percentile rank of approximately \(96.41\%\).

- Translate the requested percentile into a cumulative area to the left. - Use inverse normal technology to find the standardized cutoff. - Convert with \(x=\mu+z\sigma\) and round only the final original-scale value.

1. Use inverse normal technology with left-tail area \(0.90\) to obtain \(z\approx1.2816\). 2. Convert back: \(x=100+1.2816\cdot15\approx119.2\).

The \(90\)th percentile is approximately \(119.2\).

- Convert the middle percentage into equal left- and right-tail probabilities. - Use inverse normal technology for one standardized boundary and symmetry for the other. - Convert both boundaries to the original scale and check that the interval is centered at the mean.

1. The middle \(80\%\) leaves \(10\%\) in each tail. 2. Inverse normal technology gives standardized boundaries \(z\approx-1.2816\) and \(z\approx1.2816\). 3. The original-scale boundaries are approximately \(55.0\) and \(73.0\).

Approximately \([55.0, 73.0]\).

- Translate “lowest \(5\%\)” into a left-tail cumulative probability. - Use inverse normal technology and convert the standardized cutoff to the score scale. - Apply the same tail proportion to the group size to estimate the count.

1. Inverse normal technology for left-tail area \(0.05\) gives \(z\approx-1.6449\). 2. The cutoff is \(50+(-1.6449)\cdot8\approx36.8\). 3. The expected count is \(0.05\cdot800=40\).

Cutoff: approximately \(36.8\). Expected count: about \(40\).

- Standardize the endpoints and notice that they are opposites. - Use one normal CDF value together with symmetry to obtain the central area. - Keep full precision until rounding the final probability.

1. The endpoints are equally distant from the mean and correspond to \(z=-1.5\) and \(z=1.5\). 2. By symmetry, the central area is \(2\Phi(1.5)-1=0.8664\).

\(P(7<X<13)\approx 0.8664\).

- Translate “highest \(1\%\)” into the cumulative area below the threshold. - Use inverse normal technology to find the standardized boundary. - Convert back to the original scale and round only at the end.

1. The threshold has left-tail area \(0.99\). 2. Inverse normal technology gives \(z\approx2.3263\). 3. The threshold is \(250+2.3263\cdot30\approx319.8\).

Approximately \(319.8\).

- Keep the raw value and spread fixed while treating the center as unknown. - Translate the desired standardized position into a raw distance from the new mean. - Check that moving the mean closer to the value reduces its positive standard score.

1. With the new mean \(\mu\), the standard-score condition is \(\frac{66-\mu}{8}=1.25\). 2. Therefore, \(66-\mu=10\). 3. Solving gives \(\mu=56\).

The new mean must be \(56\).

- Both intervals have the same numerical width. - Compare their locations relative to the normal curve's peak. - Calculate each area from standard normal cumulative probabilities.

1. \(P(-1\leq Z\leq1)=\Phi(1)-\Phi(-1)\approx0.6827\). 2. \(P(0\leq Z\leq2)=\Phi(2)-\Phi(0)\approx0.4772\). 3. The first interval contains about \(0.2054\) more probability. 4. A normal density is highest at the mean and decreases as distance from the mean increases. Among these equal-width intervals, the centered interval covers more of the high-density region.

\(P(-1\leq Z\leq1)\approx0.6827\), while \(P(0\leq Z\leq2)\approx0.4772\). The centered interval contains about \(0.2054\) more probability.

- Determine where quartiles lie on the standardized normal scale. - Convert both quartile positions to the original scale. - Use normal symmetry as an additional check around the mean.

1. Normal quartiles occur at standard scores approximately \(-0.6745\) and \(0.6745\). 2. The modeled first quartile is \(50-0.6745(8)\approx44.60\). 3. The modeled third quartile is \(50+0.6745(8)\approx55.40\). 4. The reported values \(43\) and \(59\) are neither the modeled quartiles nor symmetric about the mean.

The report is inconsistent with the stated normal model. The modeled quartiles are approximately \(44.60\) and \(55.40\), not \(43\) and \(59\).

- Translate a rounded recorded value into the interval of actual measurements that produce it. - Standardize both interval endpoints. - Find the normal area between those two positions.

1. A recorded value of \(72\) represents actual values in \([71.5,72.5)\). 2. The standardized endpoints are \(z_1=\frac{71.5-70}{3}=0.5\) and \(z_2=\frac{72.5-70}{3}\approx0.8333\). 3. The probability is \(\Phi(0.8333)-\Phi(0.5)\approx0.1062\).

The probability that the recorded value is \(72\) is approximately \(0.1062\), or \(10.62\%\).

- Carry the standardized value at full precision for one calculation. - Repeat the cumulative-area calculation after rounding the standardized value. - Compare the two percentile percentages by subtraction.

1. The unrounded standard score is \(z=\frac{117-100}{15}\approx1.1333\), giving cumulative area approximately \(0.8715\). 2. Rounding first gives \(z=1.1\), with cumulative area approximately \(0.8643\). 3. Using unrounded cumulative areas, the difference is approximately \(0.00713\), or about \(0.71\) percentage points.

The percentile ranks are approximately \(87.15\%\) using the unrounded standard score and \(86.43\%\) after rounding \(z\) first. The error is approximately \(0.71\) percentage points.

- Use the boundary between possible and impossible values as the cutoff. - Standardize that boundary under the proposed model. - Evaluate whether the impossible-tail probability is negligible.

1. Standardize zero: \(z=\frac{0-3}{2}=-1.5\). 2. The model assigns \(P(X<0)=P(Z<-1.5)\approx0.0668\). 3. About \(6.68\%\) of the modeled area lies on impossible negative counts, so the normal model may be unsuitable, especially in the lower tail.

The model assigns approximately \(0.0668\), or \(6.68\%\), to impossible negative values.

- A complete standardized population must satisfy two summary conditions. - Check the average before checking the average squared distance from zero. - Passing the zero-mean condition alone is not sufficient.

1. The mean of the listed standard scores is \(\frac{-1-0.5+0.5+1}{4}=0\), which satisfies one required property. 2. Their population variance is \(\frac{(-1)^2+(-0.5)^2+(0.5)^2+1^2}{4}=\frac{2.5}{4}=0.625\). 3. Their population standard deviation is \(\sqrt{0.625}\approx0.791\), not \(1\). 4. A complete population standardized with its own population standard deviation must have mean \(0\) and population standard deviation \(1\). Therefore the report is impossible as stated.

The set is not possible as the complete population of z-scores. Its mean is \(0\), but its population standard deviation is \(\sqrt{0.625}\approx0.791\), not \(1\).

- Convert both specification limits to standardized positions. - Separate the distribution into the middle acceptable region and two failure tails. - Check that the three resulting proportions sum to approximately \(1\).

1. The standardized limits are \(z_L=\frac{170-200}{30}=-1\) and \(z_U=\frac{245-200}{30}=1.5\). 2. The acceptable proportion is \(\Phi(1.5)-\Phi(-1)\approx0.7745\). 3. The proportion below the lower limit is \(\Phi(-1)\approx0.1587\). 4. The proportion above the upper limit is \(1-\Phi(1.5)\approx0.0668\).

Acceptable: approximately \(77.45\%\). Below \(170\): approximately \(15.87\%\). Above \(245\): approximately \(6.68\%\).

- Standardize the same raw threshold separately for each machine. - Use an upper-tail area for each comparison. - Compare the resulting proportions on the same percentage scale.

1. For Machine A, \(z_A=\frac{60-50}{5}=2\), so \(P_A(X>60)\approx0.0228\). 2. For Machine B, \(z_B=\frac{60-55}{10}=0.5\), so \(P_B(X>60)\approx0.3085\). 3. Using unrounded tail probabilities, the difference is approximately \(0.2858\).

Machine A: approximately \(2.28\%\) above \(60\). Machine B: approximately \(30.85\%\) above \(60\). Machine B's proportion is about \(28.58\) percentage points greater.

- Use the symmetry of a normal distribution around its mean. - Ask where the center must lie between cutoffs with matching opposite-tail areas. - Separate the role of the center from the role of the spread.

1. Equal probabilities in opposite tails place the two cutoffs symmetrically about the mean. 2. The mean is the midpoint of the cutoffs: \(\mu=\frac{42+68}{2}=55\). 3. The standard deviation controls how large the equal tail probabilities are, but symmetry alone determines their center.

The mean is \(55\). The standard deviation is unnecessary because equal opposite-tail probabilities identify symmetric cutoffs.

- A shifted-and-scaled conversion affects the center and spread differently. - Convert both interval endpoints using the same temperature rule. - Standardize in each unit system to compare relative positions.

1. Convert the mean: \(1.8\cdot20+32=68\,{}^\circ\text{F}\). 2. Convert the standard deviation by the scale factor only: \(1.8\cdot4=7.2\,{}^\circ\text{F}\). 3. Convert the endpoints: \(14\,{}^\circ\text{C}=57.2\,{}^\circ\text{F}\) and \(27\,{}^\circ\text{C}=80.6\,{}^\circ\text{F}\). 4. In Celsius, the endpoint standard scores are \(\frac{14-20}{4}=-1.5\) and \(\frac{27-20}{4}=1.75\). 5. In Fahrenheit, they are \(\frac{57.2-68}{7.2}=-1.5\) and \(\frac{80.6-68}{7.2}=1.75\).

The Fahrenheit model has mean \(68\,{}^\circ\text{F}\) and standard deviation \(7.2\,{}^\circ\text{F}\). The safety interval is \([57.2,80.6]\,{}^\circ\text{F}\), and its endpoint standard scores remain \(-1.5\) and \(1.75\).

- First find the value where the distances to the two targets are equal. - Determine which side of that boundary favors the lower target. - Convert the resulting boundary to a standard score before finding area.

1. A value is equally distant from \(60\) and \(80\) at their midpoint, \(70\). 2. Values below \(70\) are closer to \(60\), while values above \(70\) are closer to \(80\). 3. Standardize \(70\): \(z=\frac{70-65}{10}=0.5\). 4. Therefore, \(P(X<70)=P(Z<0.5)\approx0.6915\).

The probability is approximately \(0.6915\), or \(69.15\%\).

- Translate the distance from the mean into a central standardized interval. - Find the exact normal area between the two endpoints. - Compare percentages by subtraction, not by taking a ratio.

1. The exact normal probability is \(P(-2<Z<2)\). 2. Using standard normal areas, \(P(-2<Z<2)\approx0.9545\), or \(95.45\%\). 3. The difference from the empirical-rule estimate is \(95.45\%-95.00\%=0.45\) percentage points.

The exact probability is approximately \(95.45\%\), which is about \(0.45\) percentage points higher than the empirical-rule estimate.

- Use the difference between two standardized positions to recover the scale. - Write one z-score equation for each given x-value. - Subtract the equations to eliminate the mean, then back-substitute to recover the center.

1. The x-value difference \(24\) represents \(3\) standard deviations. 2. \(\sigma=\frac{24}{3}=8\). 3. Using \(54=\mu-\sigma\), \(\mu=54+8=62\).

\(\mu=62\) and \(\sigma=8\).

- Equal percentile positions correspond to the same location on the standardized normal scale. - Express the unknown raw value relative to each distribution's center and spread. - After finding the value, use either distribution to identify its percentile.

1. Equal percentile ranks in normal distributions require equal standard scores. 2. Set \(\frac{x-60}{10}=\frac{x-70}{5}\). 3. Solving gives \(x=80\). 4. The common standard score is \(z=\frac{80-60}{10}=2\), whose percentile rank is approximately \(97.72\%\).

The raw value is \(80\), with a percentile rank of approximately \(97.72\%\) in both distributions.

- Translate the exceedance target into a percentile location for the cutoff. - Place the cutoff the appropriate standardized distance from the unknown mean. - Check that the mean must lie above the cutoff for most observations to exceed it.

1. If \(90\%\) exceed \(70\), then \(70\) is the \(10\)th percentile. 2. The standard normal value for the \(10\)th percentile is \(z\approx-1.2816\). 3. Use \(70=\mu-1.2816(4)\). 4. Solving gives \(\mu\approx75.13\).

The required mean is approximately \(75.13\).

- Use the symmetry of the required interval around the mean. - Identify the standardized half-width associated with the required central area. - A larger spread would place more probability outside the fixed interval.

1. The interval is symmetric and extends \(15\) units from the mean. 2. A central area of \(0.90\) has standard-normal cutoffs approximately \(\pm1.6449\). 3. At the greatest allowable spread, \(\frac{15}{\sigma}=1.6449\). 4. Thus \(\sigma=\frac{15}{1.6449}\approx9.12\).

The greatest allowable standard deviation is approximately \(9.12\).

- Identify the new reference group created by the given condition. - Express the conditional probability as a ratio of nested tail areas. - Use symmetry to find the area above the mean.

1. The event \(X>80\) is contained in the event \(X>70\), so \(P(X>80\mid X>70)=\frac{P(X>80)}{P(X>70)}\). 2. The standard score for \(80\) is \(z=\frac{80-70}{10}=1\), so \(P(X>80)=P(Z>1)\approx0.1587\). 3. By symmetry, \(P(X>70)=0.5\). 4. Using the unrounded tail probability, the conditional probability is approximately \(\frac{0.158655}{0.5}\approx0.3173\).

\(P(X>80\mid X>70)\approx0.3173\), or about \(31.73\%\).

- Use the symmetry of the two percentile locations to find the center. - Determine how many standard deviations separate the center from either percentile. - Relate that standardized distance to the corresponding raw-value distance.

1. The \(10\)th and \(90\)th percentiles are symmetric about the mean, so \(\mu=\frac{40+60}{2}=50\). 2. The standard normal value for the \(90\)th percentile is \(z\approx1.2816\). 3. Use \(60=50+1.2816\sigma\), giving \(\sigma=\frac{10}{1.2816}\approx7.80\).

The mean is \(50\), and the standard deviation is approximately \(7.80\).

- Evaluate the lower-tail proportion for each permitted whole-number cutoff. - Compare each modeled proportion with the maximum allowed rate. - Choose the larger cutoff only if it still satisfies the constraint.

1. At \(248\,\text{g}\), \(z=\frac{248-300}{25}=-2.08\), so \(P(X<248)\approx0.0188\). 2. At \(249\,\text{g}\), \(z=\frac{249-300}{25}=-2.04\), so \(P(X<249)\approx0.0207\). 3. The first probability is at most \(2\%\), but the second exceeds \(2\%\).

Use \(248\,\text{g}\). It classifies approximately \(1.88\%\) as underweight, while \(249\,\text{g}\) would classify approximately \(2.07\%\).

- Use equal density to locate the center between the two values. - Divide the combined probability between the symmetric tails. - Relate the raw distance from the center to the standardized tail cutoff.

1. Equal density at the two values means they are symmetric about the mean, so \(\mu=\frac{72+108}{2}=90\). 2. Symmetry splits the combined tail probability equally, giving \(0.0668\) in each tail. 3. A one-tail probability of \(0.0668\) corresponds to \(|z|\approx1.5\). 4. Since each value is \(18\) units from the mean, \(\sigma=\frac{18}{1.5}=12\).

The mean is \(90\), and the standard deviation is approximately \(12\).

- Determine how the transformation changes the center, spread, and order of values. - Standardize each value within its own distribution. - Think about what subtracting from a constant does to rankings from low to high.

1. For \(X=65\), \(z_X=\frac{65-50}{10}=1.5\), so its percentile rank is approximately \(93.32\%\). 2. The transformation gives \(\mu_Y=100-50=50\) and \(\sigma_Y=10\). 3. For \(Y=35\), \(z_Y=\frac{35-50}{10}=-1.5\), so its percentile rank is approximately \(6.68\%\). 4. The transformation reverses order: larger \(X\)-values become smaller \(Y\)-values. Therefore, corresponding percentile ranks are complementary.

For \(X=65\), the standard score is \(1.5\) and the percentile rank is approximately \(93.32\%\). For \(Y=35\), the standard score is \(-1.5\) and the percentile rank is approximately \(6.68\%\). The transformation reverses the ordering, so the two percentile ranks add to \(100\%\).