Question

A population has a mean of and a standard deviation of . suppose a simple random sample of size is selected and is used to estimate . use z-table. a. what is the probability that the sample mean will be within of the population mean (to decimals)? b. what is the probability that the sample mean will be within of the population mean (to decimals)?

Answer 1

The probability of the sample mean in each of the given options are respectively; 0.8884 and 0.9934

What is the Probability?

We are given from online complete question;

Mean; µ = 300

Standard deviation; σ = 70

Sample size; n = 125

A) For the sample to be within 8 units of the population mean, we would have sample values of 292 and 308. Thus, the expression is;

P(292 < x < 308).

z-score here is;

z = (292 - 300)/(70/√125) or z = (308 - 300)/(70/√125)

z = -1.28 or 1.28

From online p-value from z-score calculator, the p-value is 0.8884.

B) For the probability of the sample mean to be within 17 units of the population mean, we need to find the values from 283 to 317.  We want to find P(283 < x < 317).

The z-scores here are;

z = (283 - 300)/(70/√125) or z = (317 - 300)/(70/√125)

z = -2.72 or 2.72

From online p-value from z-score calculator, the p-value is 0.9934

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