Question

The proportion of Dole bananas that get damaged during shipping is thought to be 0.10. To check this, a random sample of 100 bananas is selected and 14 bananas are found to be damaged. What is the probability of having 14 or more damaged bananas in the sample?

a. What is the population proportion, or p?  

b. What is the sample proportion, or p-bar?  

c. What is the z-value of the sample proportion in part b? ROUND TO TWO (2) DECIMAL PLACES   

d. What is the probability of having 14 or more damaged bananas in the sample? Use Excel or StatCrunch to solve. ROUND TO FOUR (4) DECIMAL PLACES

Answer 1

Solution:

a. What is the population proportion, or p?  

From given information, we have population proportion, or p = 0.10.

b. What is the sample proportion, or p-bar?  

We are given

Sample size = n = 100

Number of favourable items = x = 14

Sample proportion or p-bar = x/n = 14/100 = 0.14

p̄ = 0.14

c. What is the z-value of the sample proportion in part b?

The formula for z-value is given as below:

Z = (p̄ - p) / sqrt(pq/n)

Where, q = 1 – p = 1 – 0.10 = 0.90

Z = (0.14 – 0.10) / sqrt(0.10*0.90/100)

Z = 0.04/ 0.03

Z = 1.333333

Z = 1.33

d. What is the probability of having 14 or more damaged bananas in the sample?

We have to find P(X≥14) = P(p̄≥0.14)

P(p̄ ≥ 0.14) = 1 – P(p̄ < 0.14)

Z = (p̄ - p) / sqrt(pq/n)

Z = (0.14 – 0.10) / sqrt(0.10*0.90/100)

Z = 0.04/ 0.03

Z = 1.333333

P(Z<1.333333) = P(p̄ < 0.14) = 0.908789

(by using excel)

[Use excel command =NORMSDIST(1.333333) and ENTER]

P(p̄ ≥ 0.14) = 1 – P(p̄ < 0.14)

P(p̄ ≥ 0.14) = 1 – 0.908789

P(p̄ ≥ 0.14) = 0.091211

P(X≥14) = 0.091211

Required probability = 0.0912