Question

Majesty Video Production Inc. wants the mean length of its advertisements to be 29 seconds. Assume the distribution of ad length follows the normal distribution with a population standard deviation of 2 seconds. Suppose we select a sample of 17 ads produced by Majesty.

Required:
a. What can we say about the shape of the distribution of the sample mean time?
b. What is the standard error of the mean time?
c. What percent of the sample means will be greater than 30.05 seconds?
d. What percent of the sample means will be greater than 27.75 seconds?
e. What percent of the sample means will be greater than 27.75 but less than 30.05 seconds?

Answer 1

a) By the Central limit theorem, approximately normally distributed, with mean of 29 seconds and standard deviation of 0.4851 seconds.

b) 0.4851 seconds

c) 1.54% of the sample means will be greater than 30.05 seconds

d) 99.51% of the sample means will be greater than 27.75 seconds

e) 97.97% of the sample means will be greater than 27.75 but less than 30.05 seconds

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.

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(standard error of the mean)
`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 problem, we have that:

`\mu = 29, \sigma = 2, n = 17, s = (2)/(√(17)) = 0.4851`

a. What can we say about the shape of the distribution of the sample mean time?

By the Central limit theorem, approximately normally distributed, with mean of 29 seconds and standard deviation of 0.4851 seconds.

b. What is the standard error of the mean time?

0.4851 seconds.

c. What percent of the sample means will be greater than 30.05 seconds?

This is 1 subtracted by the pvalue of Z when X = 30.05. So

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

By the Central Limit Theorem

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

`Z = (30.05 - 29)/(0.4851)`

`Z = 2.16`

`Z = 2.16` has a pvalue of 0.9846.

1 - 0.9846 = 0.0154

1.54% of the sample means will be greater than 30.05 seconds

d. What percent of the sample means will be greater than 27.75 seconds?

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

`Z = (27.75 - 29)/(0.4851)`

`Z = -2.58`

`Z = -2.58` has a pvalue of 0.0049

1 - 0.0049 = 0.9951

99.51% of the sample means will be greater than 27.75 seconds

e. What percent of the sample means will be greater than 27.75 but less than 30.05 seconds?

This is the pvalue of Z when X = 30.05. So

X = 30.05

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

`Z = (30.05 - 29)/(0.4851)`

`Z = 2.16`

`Z = 2.16` has a pvalue of 0.9846.

X = 27.75

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

`Z = (27.75 - 29)/(0.4851)`

`Z = -2.58`

`Z = -2.58` has a pvalue of 0.0049

0.9846 - 0.0049 = 0.9797

97.97% of the sample means will be greater than 27.75 but less than 30.05 seconds