3. The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population proportion is p = .17 and a sample of 800 households will be selected from the population. a. Show the sampling distribution of p, the sample proportion of households spending more than $100 per week on groceries. b. What is the probability that the sample proportion will be within ±.02 of the population proportion? c. Answer part (b) for a sample of 1600 household

See the steps below.

Step-by-step explanation:

a. Show the sampling distribution of p, the sample proportion of households spending more than $100 per week on groceries.  

The sample means = 0.17

The standard deviation of the sample means = = 0.0133

Therefore, the sample proportion of households spending more than $100 per week on groceries is approximately 0.0133.

b) What is the probability that the sample proportion will be within plus or minus .02 of the population proportion?

Since the sample means = 0.17

For plus 0.02

The sample means = 0.17 + 0.02 = 0.19

Therefore, we have:

z(0.19) = = 1.5060

For minus 0.02

The sample means = 0.17 - 0.02 = 0.15

z(0.15) = = - 1.5060

Therefore, we now have:

P(0.15< p < 0.19) = P(-1.5060< z <1.5060) = 0.8679

c) Answer part (b) for a sample of 1600.

For plus 0.02

The sample means = 0.17 + 0.02 = 0.19

Therefore, we have:

z(0.19) = = 2.1297

For minus 0.02

The sample means = 0.17 - 0.02 = 0.15

z(0.15) = = - 2.1297

Therefore, we now have:

P(0.15< p < 0.19) = P(-2.197< z <2.1297) = 0.9660

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b