Question

In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weighs more than 180 pounds? Assume that we want 96% confidence that the error is no more than 2 percentage points.

A) 923
B) 1267
C) 2628
D) 2001

Answer 1

B) 1267

Explanation:

Percentage of people above 180 pounds.

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

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

`Z = (X - \mu)/(\sigma)`

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:

`\mu = 152, \sigma = 26`

This percentage is 1 subtracted by the pvalue of Z when X = 26. So

`Z = (X - \mu)/(\sigma)`

`Z = (180 - 152)/(26)`

`Z = 1.08`

`Z = 1.08` has a pvalue of 0.8599

1 - 0.8599 = 0.1401

14.01% of people above 180 pounds.

Confidence interval for the proportion:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is given by:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

For this question, we have that:

`\pi = 0.1401`

96% confidence level

So
`\alpha = 0.04`, z is the value of Z that has a pvalue of
`1 - (0.04)/(2) = 0.98`, so
`Z = 2.054`.

We need

A sample size of n.

n is found when M = 0.02. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.02 = 2.054\sqrt{(0.1401*0.8599)/(n)}`

`0.02√(n) = 2.054√(0.1401*0.8599)`

`√(n) = (2.054√(0.1401*0.8599))/(0.02)`

`(√(n))^(2) = ((2.054√(0.1401*0.8599))/(0.02)}^(2)`

`n \cong 1267`

So the correct answer is:

B) 1267