Question

College students with a checking account typically write relatively few checks in any given month, whereas nonstudent residents typically write many more checks during a month. Suppose that \(50 \%\) of a bank's accounts are held by students and that \(50 \%\) are held by nonstudent residents. Let \(x\) denote the number of checks written in a given month by a randomly selected bank customer. a. Give a sketch of what the probability distribution of \(x\) might look like. b. Suppose that the mean value of \(x\) is \(22.0\) and that the standard deviation is 16.5. If a random sample of \(n=100\) customers is to be selected and \(\bar{x}\) denotes the sample average number of checks written during a particular month, where is the sampling distribution of \(\bar{x}\) centered, and what is the standard deviation of the \(\bar{x}\) distribution? Sketch a rough picture of the sampling distribution. c. Referring to Part (b), what is the approximate probability that \(\bar{x}\) is at most \(20 ?\) at least 25 ?

Short answer

Part (a) : The distribution diagram would show bimodal character with peaks at low and high check counts. Part (b) : The sampling distribution of \(\bar{x}\) is centered at the mean of the population, which is 22.0 and has a standard deviation of 1.65. Part (c) : The approximate probabilities that \(\bar{x}\) is at most 20 and at least 25 would be found by calculating the z-scores and referring to the z-table or using statistical software.

Step-by-step solution

Part (a): Create a rough sketch of the distribution diagram

Given that college students typically write fewer checks and nonstudent residents write more, the shape of the distribution would possibly be bimodal, with a peak at low check counts which would be students, and a peak at high check counts which would be non-student residents.

Part (b): Find the mean and standard deviation of the sampling distribution

According to the Central Limit Theorem, the sampling distribution of \(\bar{x}\) is normally distributed with mean equal to the population mean and standard deviation equals to the population standard deviation divided by the square root of the sample size. Consequently, the mean of \(\bar{x}\), is the same as the mean of the population (\(\mu\)), which is 22.0. The standard deviation of the sampling distribution of \(\bar{x}\), is the standard deviation of the population (\(\sigma\)) divided by the square root of the sample size (\(n\)), which is \(16.5 / \sqrt{100} = 1.65\). A rough sketch of the sampling distribution would be a normal distribution centered at 22.

Part (c): Calculate probability of \(\bar{x}\) being within certain limits

To calculate the probability that \(\bar{x}\) is at most 20 or at least 25, we first calculate the z-scores for these values. The z-score is calculated as \((value - mean) / standard deviation\). For 20, it would be \((20 - 22) / 1.65 = -1.21\), and for 25, it would be \((25 - 22) / 1.65 = 1.82\). We then use the z-table or statistical software to find the respective probabilities. Probability that \( \bar{x} \leq 20 \) is P(Z < -1.21) and the probability that \( \bar{x} \geq 25 \) is P(Z > 1.82).