Question

The times that customers spend in a book store are normally distributed with a mean of 39.5 minutes and a standard deviation of 15.9 minutes. A random sample of 60 customers has a mean of 36.1 minutes or less. Would this outcome be considered unusual, so that the store should reconsider its displays?

Answer 1

Using standard error and z-score calculations, the sample mean time of 36.1 minutes for customers in the bookstore is within the expected range for a normal distribution and is not considered unusual. Therefore, the store may not need to make adjustments based on this outcome alone.

Step-by-step explanation:

To determine if the mean time of 36.1 minutes for the 60 customers is unusual, we need to find the standard error of the mean and calculate the z-score. The standard error (SE) is given by the formula SE = σ/√n, where σ is the population standard deviation and n is the sample size. Plugging in the numbers, we get SE = 15.9/√60 = 2.0513. The z-score is calculated as z = (X - μ) / SE, where X is the sample mean, and μ is the population mean. For our problem, z = (36.1 - 39.5) / 2.0513 = -1.6565.

To decide whether this z-score is unusual, we typically look at the standard normal distribution. A z-score below -2 or above 2 is usually considered unusual in a standard normal distribution, which correlates to the tails of the curve, representing the outer 5% of the data. However, with a z-score of approximately -1.66, this would not be considered highly unusual since it's within the -2 to 2 range. Therefore, the store might not need to reconsider its displays based solely on this data.

Answer 2

Since |Z| = 1.66 < 2, this outcome should not be considered unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If |Z| > 2, X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

`\mu = 39.5, \sigma = 15.9, n = 60, s = (15.9)/(√(60)) = 2.05`

A random sample of 60 customers has a mean of 36.1 minutes or less. Would this outcome be considered unusual, so that the store should reconsider its displays?

We have to find Z when X = 36.1.

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (36.1 - 39.5)/(2.05)`

`Z = -1.66`

So |Z| = 1.66

Since |Z| = 1.66 < 2, this outcome should not be considered unusual.