According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

Let p = the proportion of keypads that pass inspection at a cell phone assembly plant.

The probability that a randomly selected cell phone keypad passes the inspection is, p = 0.77.

A random sample of n = 111 keypads is analyzed.

Then the sampling distribution of $\hat p$ is:

$\hat p\sim N(p,\ \sqrt{\frac{p(1-p)}{n}})\\\Rightarrow \hat p\sim N (0.77, 0.04)$

Compute the probability that the proportion of passed keypads is between 0.72 and 0.80 as follows:

$P(0.72$

$=P(-1.25$

Thus, the probability that the proportion of passed keypads is between 0.72 and 0.80 is 0.6677.

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3 years ago