Question

Weights of women in one age group are normally distributed with a standard deviation sigma of 20 lb. A researcher wishes to estimate the mean weight of all women in this age group. Find how large a sample must be drawn in order to be​ 90% confident that the sample mean will not differ from the population mean by more than 3.5 lb.

Answer 1

Answer:

88.344 or Rounding up to 89

Step-by-step explanation:

When talking about the confidence level, we need to find the z-score for 90% confidence. Since z-scores are constant values we can easily find these online. In this case the z-score for 90% confidence is z = 1.64485

The formula for z-score is as follow

$z = \frac{Sample Mean  - Population Mean}{a}$

Where

$a = \frac{Standard Deviation}{Sample Size}$

Now we know from the question that

$Sample Mean - Population Mean = 3.5$

We can easily solve the above equation to find the Standard Deviation

$1.64485 = \frac{3.5}{a}\\ a =2.12785$

To find Population Size

$a = \frac{Standard Deviation}{\sqrt{Sample Size} }\\ 2.12785 = \frac{20}{\sqrt{Sample Size}} \\ Sample Size = 88.344$

Answer 2

The sample size where the sample mean will not differ from the population mean by more than 3.5 lb is; 88

What is the required sample size?

The formula for margin of error is;

E = z(σ/√n)

where;

z is sample statistic at confidence level

σ is standard deviation

n is sample size

Since we want to find the sample size where the sample mean will not differ from the population mean by more than 3.5 lb. Thus;

E = 3.5

we have;σ = 20 lb

z at 90% CL = 1.645

Thus;

3.5 = 1.645(20/√n)

n = (1.645 * 20/3.5)²

n = 88.36

n ≈ 88

Read more about Sample Size at; https://brainly.com/question/14470673