Question

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes. Suppose 64 visitors independently view the site. Determine the following: a. The expected value and the variance of the mean time of the visitors. b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes c. The value exceeded by the mean time of the visitors with probability 0.01.

Answer 1

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$.

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

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.25 - 10}{0.25}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

X = 9.75

$Z = \frac{X - \mu}{s}$

$Z = \frac{9.75 - 10}{0.25}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.8413 - 0.1587 = 0.6826.

0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c. The value exceeded by the mean time of the visitors with probability 0.01.

The 100 - 1 = 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.

$Z = \frac{X - \mu}{s}$

$2.327 = \frac{X - 10}{0.25}$

$X - 10 = 2.327*0.25$

$X = 10.58$

So 10.58 minutes.