Question

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

a. What is the probability that the sample proportion is greater than 0.63?

b. What is the probability that the sample proportion is between 0.60 and 0.66?

c. What is the probability that the sample proportion is greater than 0.592

d. What is the probability that the sample proportion is between 0.57 and 0.60?

e. What is the probability that the sample proportion is less than 0.51?

Answer 1

Explained below.

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{(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{(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((\hat p-\mu_(\hat p))/(\sigma_(\hat p))>(0.63-0.61)/(0.019))\\\\=P(Z>1.05)\\\\=1-P(Z<1.05)\\\\=1-0.85314\\\\=0.14686\\\\\approx 0.1469`

(b)

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

`P(0.60<\hat p<0.66)=P((0.60-0.61)/(0.019)<(\hat p-\mu_(\hat p))/(\sigma_(\hat p))<(0.66-0.61)/(0.019))\\\\=P(-0.53<Z<2.63)\\\\=P(Z<2.63)-P(Z<-0.53)\\\\=0.99573-0.29806\\\\=0.69767\\\\\approx 0.6977`

(c)

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

`P(\hat p>0.592)=P((\hat p-\mu_(\hat p))/(\sigma_(\hat p))>(0.592-0.61)/(0.019))\\\\=P(Z>-0.95)\\\\=P(Z<0.95)\\\\=0.82894\\\\\approx 0.8289`

(d)

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

`P(0.57<\hat p<0.60)=P((0.57-0.61)/(0.019)<(\hat p-\mu_(\hat p))/(\sigma_(\hat p))<(0.60-0.61)/(0.019))\\\\=P(-2.11<Z<-0.53)\\\\=P(Z<-0.53)-P(Z<-2.11)\\\\=0.29806-0.01743\\\\=0.28063\\\\\approx 0.2806`

(e)

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

`P(\hat p<0.51)=P((\hat p-\mu_(\hat p))/(\sigma_(\hat p))<(0.51-0.61)/(0.019))\\\\=P(Z<-5.26)\\\\=0`