Question

A population proportion is 0.61. Suppose a random sample of 658 items is sampled randomly from this population

Answer 1

Answer:

Explained below.

Step-by-step explanation:

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:

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

A random sample of n = 658 items is sampled randomly from this population.

As the sample size is large, i.e. n = 658 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample proportion by a Normal distribution.

Compute the mean and standard deviation as follows:

$\mu_{\hat p}=0.61\\\\\sigma_{\hat p}=\sqrt{\frac{0.61(1-0.61)}{658}}=0.019$

(a)

Compute the probability that the sample proportion is greater than 0.63 as follows:

$P(\hat p>0.63)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}>\frac{0.63-0.61}{0.019})\\\\=P(Z>1.05)\\\\=1-P(Z$

(b)

Compute the probability that the sample proportion is between 0.60 and 0.66 as follows:

$P(0.60$

(c)

Compute the probability that the sample proportion is greater than 0.592 as follows:

$P(\hat p>0.592)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}>\frac{0.592-0.61}{0.019})\\\\=P(Z>-0.95)\\\\=P(Z$

(d)

Compute the probability that the sample proportion is between 0.57 and 0.60 as follows:

$P(0.57$

(e)

Compute the probability that the sample proportion is less than 0.51 as follows:

$P(\hat p$