Suppose a large shipment of microwave ovens contained 7% defectives. If a sample of size 300 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 3%

95.86% probability that the sample proportion will differ from the population proportion by less than 3%

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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.

In this problem, we have that:

For a proportion, we have that:

What is the probability that the sample proportion will differ from the population proportion by less than 3%

This is the pvalue of Z when X = 0.07 + 0.03 = 0.1 subtracted by the pvalue of Z when X = 0.07 - 0.03 = 0.04. So

X = 0.1

has a pvalue of 0.9793

X = 0.04

has a pvalue of 0.0207

0.9793 - 0.0207 = 0.9586

95.86% probability that the sample proportion will differ from the population proportion by less than 3%

answer: no

step-by-step explanation:

this is the formula: v = the value of investment at the end of the time period

p = the principal amount (the initial amount invested)

r = the annual interest rate

n = the annual frequency of compounding (how many times a year interest is added)

t = the number of years the money is invested

^ means raise to the power of