Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that

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Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that

a SRS of size 200 would come within plus or minus 3 percentage points of this true value. In other words, find probability that pˆ takes a value between 0.17 and 0.23.

Mathematics

1 answer:

zzz [12.3K] · Answer 1 · 2026-03-02T00:44:01+03:00

Answer:

There is a 71.08% chance that pˆ lies between 0.17 and 0.23.

Step-by-step explanation:

We apply the binomial approximation to the normal distribution for this problem.

Binomial probability distribution

This describes the likelihood of achieving exactly x successes across n trials, considering probability p.

This can be approximated using normal distribution principles with the expected value and standard deviation.

Expected value in a binomial context is:

$E(X) = np$

Standard deviation in a binomial context is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

We can solve problems with normally distributed samples via the z-score formula.

Within a data set having mean $\mu$ and standard deviation $\sigma$ , the z-score for a value X is calculated as:

$Z = \frac{X - \mu}{\sigma}$

The Z-score indicates how far the value is from the mean in standard deviations. After calculating the Z-score, we consult the z-score table to identify the corresponding p-value. This p-value reflects the probability that the measure is less than X, signifying X's percentile. To find the likelihood that the measure exceeds X, we subtract the p-value from 1.

For approximating between a binomial and normal distribution, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .