A sample of manufactured items will be selected from a large population in which 8 percent of the items are defective of the following, which is the least value for a sample size that will allow for the sampling distribution of the sample proportion to be assumed approximately normal?

11
30
100
125
250

According to the Central limit theorem, if from an unknown population large samples of sizes n ≥ 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

[tex]\mu_{\hat p}=p[/tex]

The standard deviation of this sampling distribution of sample proportion is:

[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]

Here it is provided that manufactured items are selected from a large population in which 8% of the items are defective.

That is, the proportion of defective items is, p = 0.08.

For the sampling distribution of the sample proportion of defective items to be normally distributed the least sample size required is 30.

Thus, the sample size required should be at least of size 30.

trinitysterling1227 trinitysterling1227 · Answer 2 · 2021-01-25T05:10:42+01:00

Answer:

the correct answer is 125

Step-by-step explanation:

The value of n must be large enough so that both np≥10 and n(1−p)≥10. If n=125, then np=(125)(0.08)=10 and n(1−p)=(125)(1−0.08)=115, both of which are greater than or equal to 10. Any value less than 125 will result in np having a value less than 10, so 125 is the least value of n to ensure that the sampling distribution of the sample proportion is approximately normal.