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Joint, marginal, and conditional relative frequency

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53083912
A simplified survival table shows how many men from an initial group of \(100{,}000\) live births reached selected ages in two different years. <table> <tr><th>Age in years</th><th>\(1960\)</th><th>\(2020\)</th></tr> <tr><td>\(0\)</td><td>\(100{,}000\)</td><td>\(100{,}000\)</td></tr> <tr><td>\(40\)</td><td>\(92{,}400\)</td><td>\(98{,}800\)</td></tr> <tr><td>\(70\)</td><td>\(52{,}100\)</td><td>\(84{,}500\)</td></tr> <tr><td>\(85\)</td><td>\(12{,}300\)</td><td>\(42{,}600\)</td></tr> </table> a) For each year, estimate the probability that a man who has reached age \(40\) will reach age \(70\). b) For each year, estimate the probability that a man who has reached age \(40\) will reach at least age \(85\). c) Explain why these are empirical probabilities rather than probabilities based on equally likely outcomes.

Hints

- For each year, use the number reaching the later age as the numerator. - Because the person has already reached age \(40\), use the age-\(40\) count as the denominator. - Distinguish probabilities estimated from data from probabilities derived from equally likely theoretical outcomes. - Identify the population represented by each column.

Solution

1. For 1960, the conditional proportion reaching age \(70\) among those reaching age \(40\) is \(\frac{52100}{92400}\approx 0.5639\). For 2020, it is \(\frac{84500}{98800}\approx 0.8553\). 2. For 1960, the conditional proportion reaching age \(85\) among those reaching age \(40\) is \(\frac{12300}{92400}\approx 0.1331\). For 2020, it is \(\frac{42600}{98800}\approx 0.4312\). 3. The values are estimated from observed relative frequencies in population data. They are not derived from a sample space in which all elementary outcomes are assumed to be equally likely.

Answer

a) 1960: \(0.5639\), or about \(56.39\%\). 2020: \(0.8553\), or about \(85.53\%\). b) 1960: \(0.1331\), or about \(13.31\%\). 2020: \(0.4312\), or about \(43.12\%\). c) They are empirical probabilities because they are based on observed relative frequencies.
53084012
An electronics company tracks the reliability of \(50{,}000\) smartphones. The table shows how many phones are still fully functional after selected numbers of months. <table> <tr><th>Months in use</th><th>Functional phones</th></tr> <tr><td>\(0\)</td><td>\(50{,}000\)</td></tr> <tr><td>\(12\)</td><td>\(48{,}500\)</td></tr> <tr><td>\(24\)</td><td>\(42{,}000\)</td></tr> <tr><td>\(36\)</td><td>\(25{,}000\)</td></tr> <tr><td>\(48\)</td><td>\(8000\)</td></tr> </table> a) Estimate the probability that a phone still working after \(12\) months is also working after \(36\) months. b) Estimate the probability that a phone working after \(24\) months fails during the third year, between months \(24\) and \(36\). c) Explain why failure time cannot be modeled here by assuming all possible months of failure are equally likely.

Hints

- In a conditional probability, use only the phones that satisfy the stated starting condition. - Find the number of failures during a time interval by subtracting the later survivor count from the earlier survivor count. - Consider whether wear makes failure equally likely at every age.

Solution

1. For part a, restrict the group to the \(48{,}500\) phones working after \(12\) months. Of those, \(25{,}000\) are working after \(36\) months, so the estimate is \(\frac{25000}{48500}\approx 0.5155\). 2. Between months \(24\) and \(36\), \(42{,}000-25{,}000=17{,}000\) phones fail. Among phones working at month \(24\), the estimated conditional probability is \(\frac{17000}{42000}\approx 0.4048\). 3. Failure rates can change with age and wear. The data do not support treating failure in each month as an equally likely outcome.

Answer

a) \(\frac{25000}{48500}\approx 0.5155\), or about \(51.55\%\). b) \(\frac{17000}{42000}\approx 0.4048\), or about \(40.48\%\). c) The probability of failure changes over time, so the possible failure months are not equally likely.
53095112
A survival table gives the number \(l_x\) of people from an initial population who reach age \(x\). <table> <tr><th>Age \(x\)</th><th>Number \(l_x\) reaching age \(x\)</th></tr> <tr><td>\(25\)</td><td>\(98{,}412\)</td></tr> <tr><td>\(45\)</td><td>\(95{,}120\)</td></tr> <tr><td>\(65\)</td><td>\(80{,}455\)</td></tr> </table> Estimate each probability. 1. A person who is currently \(25\) reaches age \(65\). 2. Two people who are currently \(45\) both reach age \(65\). Assume their survival events are independent. 3. A person who is currently \(25\) dies after reaching age \(45\) but before reaching age \(65\).

Hints

- Estimate a conditional probability by dividing the later-age count by the count for the current age. - For two independent events that must both occur, multiply their probabilities. - To count deaths between two ages, subtract the later survivor count from the earlier survivor count. - Use the population that has already reached the stated current age as the denominator.

Solution

1. Among people reaching age \(25\), the estimated proportion reaching age \(65\) is \(\frac{80455}{98412}\approx 0.8175\). 2. For one person currently age \(45\), the estimated probability of reaching age \(65\) is \(\frac{80455}{95120}\approx 0.8458\). Under the stated independence assumption, the probability that both people reach age \(65\) is \(\left(\frac{80455}{95120}\right)^2\approx 0.7154\). 3. The number reaching age \(45\) but not age \(65\) is \(95{,}120-80{,}455=14{,}665\). Relative to those reaching age \(25\), the estimated probability is \(\frac{14665}{98412}\approx 0.1490\).

Answer

1. \(0.8175\), or about \(81.75\%\). 2. \(0.7154\), or about \(71.54\%\). 3. \(0.1490\), or about \(14.90\%\).
52213012
A company operates two IT support centers, \(\alpha\) and \(\beta\). Each center handles hardware requests \((H)\) and software requests \((S)\). Let \(L\) be the event that a request is resolved. The data from the most recent quarter are shown below. <table> <tr><td>Center</td><td>Request type</td><td>Total requests</td><td>Resolved requests</td></tr> <tr><td>\(\alpha\)</td><td>Hardware \((H)\)</td><td>\(200\)</td><td>\(40\)</td></tr> <tr><td>\(\alpha\)</td><td>Software \((S)\)</td><td>\(800\)</td><td>\(720\)</td></tr> <tr><td>\(\beta\)</td><td>Hardware \((H)\)</td><td>\(800\)</td><td>\(200\)</td></tr> <tr><td>\(\beta\)</td><td>Software \((S)\)</td><td>\(200\)</td><td>\(190\)</td></tr> </table> a) For each center, find \(P(L\mid H)\) and \(P(L\mid S)\). b) A manager claims that center \(\alpha\) is much more effective because it resolved \(76\%\) of all requests, while center \(\beta\) resolved only \(39\%\). Verify these overall rates. c) Evaluate the manager's claim. Which center performs better within each request type? Explain why the type-specific rates and overall rates lead to different conclusions.

Hints

- Interpret \(P(L\mid H)\) as a rate within the hardware-request group. - Add the total and resolved requests separately for each center. - Compare the two centers within hardware, then compare them within software. - Examine the mix of hardware and software requests at each center.

Solution

1. For center \(\alpha\), \(P(L\mid H)=\frac{40}{200}=0.20\) and \(P(L\mid S)=\frac{720}{800}=0.90\). 2. For center \(\beta\), \(P(L\mid H)=\frac{200}{800}=0.25\) and \(P(L\mid S)=\frac{190}{200}=0.95\). 3. Center \(\alpha\) resolved \(40+720=760\) of \(1000\) requests, so its overall rate is \(\frac{760}{1000}=0.76\). Center \(\beta\) resolved \(200+190=390\) of \(1000\) requests, so its overall rate is \(\frac{390}{1000}=0.39\). 4. Center \(\beta\) has the higher resolution rate for both hardware and software. Its lower overall rate occurs because \(80\%\) of its requests are hardware requests, which have much lower resolution rates at both centers. Center \(\alpha\) receives mostly software requests. This reversal is an example of how aggregated data can obscure subgroup comparisons.

Answer

a) Center \(\alpha\): \(P(L\mid H)=20\%\), \(P(L\mid S)=90\%\). Center \(\beta\): \(P(L\mid H)=25\%\), \(P(L\mid S)=95\%\). b) Center \(\alpha\): \(76\%\). Center \(\beta\): \(39\%\). c) Center \(\beta\) performs better for both hardware and software. Its overall rate is lower because it handles a much larger proportion of the harder-to-resolve hardware requests.

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