Question

In a certain population, body weights are normally distributed with a mean of 152 poundsand a standard deviation of 26 pounds. How many people must be surveyed if we want toestimate the percentage who weigh more than 180 pounds

Answer 1

To estimate the percentage who weigh more than 180 pounds, we need to survey 316 people approximately.

In the question, we are given that in a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds.

We are asked to find the number of people that needs to be surveyed if we want to estimate the percentage who weigh more than 180 pounds.

The population mean given to us is, μ = 152 pounds.

The population standard deviation given to us is, σ = 26 pounds.

The value to estimate given to us is, x = 180 pounds.

The true proportion value p, corresponding to this, can be calculated as:

P(X > 180)

= 1 - P(X < 180)

= 1 - P(Z < 180-152/26)

= 1 - P(Z < 1.08)

= 1 - 0.8599

= 0.1401.

Thus, p = 0.1401.

The confidence level given to us is 96%.

The z-score corresponding to this is, z = 2.05.

The Margin of error given to us is, MOE = 4% = 0.04.

We are asked to find the sample size n.

The formula for this is:

$n = \frac{z^2p(1-p)}{MOE^2}\\= \frac{2.05^2(0.1401)(1-0.1401)}{0.04^2}\\= \frac{4.2025*0.1401*0.8599}{0.0016}\\ = \frac{0.50628353797}{0.0016}\\ = 316.427211234\\\approx 316$

Thus, to estimate the percentage who weigh more than 180 pounds, we need to survey 316 people approximately.

Learn more about estimating the sample size at

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The provided question is incomplete. The complete question is:

"In a certain population, body weights are normally distributed with a mean of 152 pounds 30) and a standard deviation of 26 pounds. how many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? assume that we want 96% confidence that the error is no more than 4 percentage points."