Using the <u>normal distribution and the central limit theorem</u>, it is found that:
a) The standard deviation is of $955.34.
b) 0.5 = 50% probability that the sample mean will be more than $27,175.
c) There is a 0.2938 = 29.38% probability that the sample mean will be within $1,000 of population mean.
d) There is a 0.177 = 17.7% probability that the sample mean will be within $1,000 of population mean.
In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- 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
.
In this problem:
- The average is of $27,175, hence
.
- The population standard deviation is of $7,400, hence
.
- A sample of 60 students is taken, hence
.
Item a:
The standard deviation is of $955.34.
Item b:
This probability is <u>1 subtracted by the p-value of Z when X = 27175</u>, hence:
By the Central Limit Theorem
has a p-value of 0.5.
1 - 0.5 = 0.5
0.5 = 50% probability that the sample mean will be more than $27,175.
Item c:
The probability is P(|Z| > 1.05), which is <u>2 multiplied by the p-value of Z = -1.05.</u>
Looking at the z-table, Z = -1.05 has a p-value of 0.1469.
2 x 0.1469 = 0.2938
There is a 0.2938 = 29.38% probability that the sample mean will be within $1,000 of population mean.
Item d:
Sample of 100, hence
Z = -1.35 has a p-value of 0.0885.
2 x 0.0885 = 0.177
There is a 0.177 = 17.7% probability that the sample mean will be within $1,000 of population mean.
To learn more about the <u>normal distribution and the central limit theorem</u>, you can take a look at brainly.com/question/24663213