Question

In 2017, a website reported that only 10% of surplus food is being recovered in the food-service and restaurant sector, leaving approximately 1.5 billion meals per year uneaten. Assume this is the true population proportion and that you plan to take a sample survey of 565 companies in the food service and restaurant sector to further investigate their behavior.

(a) Show the sampling distribution of p, the proportion of food recovered by your sample respondents.

(b) What is the probability that your survey will provide a sample proportion within ±0.03 of the population proportion? (Round your answer to four decimal places.)

(c)What is the probability that your survey will provide a sample proportion within ±0.015 of the population proportion? (Round your answer to four decimal places.)

Answer 1

a)

sample proportion =0.10

std.dev = sqrt(p *(1-p)/n)
= sqrt( 0.10 *(1-0.10)/565)
= 0.0126

b)

Here, μ = 0.1, σ = 0.0126, x1 = 0.07 and x2 = 0.13. We need to compute P(0.07<= X <= 0.13). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.07 - 0.1)/0.0126 = -2.38
z2 = (0.13 - 0.1)/0.0126 = 2.38

Therefore, we get
P(0.07 <= X <= 0.13) = P((0.13 - 0.1)/0.0126) <= z <= (0.13 - 0.1)/0.0126)
= P(-2.38 <= z <= 2.38) = P(z <= 2.38) - P(z <= -2.38)
= 0.9913 - 0.0087
= 0.9826

c)

Here, μ = 0.1, σ = 0.0126, x1 = 0.085 and x2 = 0.115. We need to compute P(0.085<= X <= 0.115). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.085 - 0.1)/0.0126 = -1.19
z2 = (0.115 - 0.1)/0.0126 = 1.19

Therefore, we get
P(0.085 <= X <= 0.115) = P((0.115 - 0.1)/0.0126) <= z <= (0.115 - 0.1)/0.0126)
= P(-1.19 <= z <= 1.19) = P(z <= 1.19) - P(z <= -1.19)
= 0.883 - 0.117
= 0.7660