Question

Jonathan, a college business student wants to study how many pages of notes fellow classmates took during a semester for a finance class. Past data reveals that pages of notes for a finance college class has a mean of 95 , with standard deviation 25 pages. He plans to take a random sample of 30 such college students and will calculate the mean pages of notes they take to compare to the known pages of notes students take for a finance college class. For these values, the mean and standard deviation of the sampling distribution of sample means for a sample of size 30 are: μx¯=95 and σx¯=2530√=4.6. What is the probability that the sample mean for a sample of size 30 will be at least 99? You may use a calculator or the portion of the z-table given below.

Answer 1

Using the normal distribution and the central limit theorem, it is found that there is a 0.1913 = 19.13% probability that the sample mean for a sample of size 30 will be at least 99.

Normal Probability Distribution

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

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

• It 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, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For the population, the mean and the standard deviation are given by, respectively, $\mu = 95$ and $\sigma = 25$.

For a sample of 30, we have that:

$n = 30, s = \frac{25}{\sqrt{30}} = 4.6$

The probability that the sample mean for a sample of size 30 will be at least 99 is one subtracted by the p-value of Z when X = 99, hence:

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

By the Central Limit Theorem

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

$Z = \frac{99 - 95}{4.6}$

$Z = 0.87$

$Z = 0.87$ has a p-value of 0.8078.

1 - 0.8087 = 0.1913.

0.1913 = 19.13% probability that the sample mean for a sample of size 30 will be at least 99.

More can be learned about the normal distribution and the central limit theorem at https://brainly.com/question/24663213