Question

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is .
n = 160, p = 0.29; P( > 0.3)

Answer 1

Answer:

P(X > 0.3) = 0.3897

Step-by-step explanation:

Normal probability distribution:

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:

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

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

Applying the central limit theorem:

$\mu = p = 0.29, s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.29*0.71}{160}} = 0.0359$

P( > 0.3)

This is 1 subtracted by the pvalue of Z when X = 0.3. So

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

By the Central Limit Theorem

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

$Z = \frac{0.3 - 0.29}{0.0359}$

$Z = 0.28$

$Z = 0.28$ has a pvalue of 0.6103

1 - 0.6103 = 0.3897

So

P(X > 0.3) = 0.3897