Question

The Apple, Inc. sales manager for the Chicago West Suburban region is disturbed about the large number of complaints her office is receiving about defective ipods. New Apple CEO Tim Cook is also very displeased and wants the situation remedied as soon as possible. The sales manager examines this Excel file ipod Weekly Complaints, which shows the number of complaints concerning defective ipods received by her office for each of the immediately preceding 52 weeks. Question 1. Make a histogram of the number of complaints received each week. As your bin range use the values 60, 80, 100, 120, 140, 160 listed in the Excel file. Then select the choice below that is the most appropriate conclusion from the histogram (2 submissions allowed). o Based on the histogram, it is appropriate to use a normal model to describe the distribution of the number of complaints received each O The histogram is severely left-skewed; therefore a normal model is not appropriate to describe the number of complaints received each week. The histogram appears to be bi-modal; therefore a normal model is not appropriate to describe the number of complaints received each week O The histogram has several outliers; therefore a normal model is not appropriate to describe the number of complaints received each week. O The histogram is severely right-skewed; therefore a normal model is not appropriate to describe the number of complaints received each week. Question 2.What is the probability that between 84 and 111 complaints are received in one week? Use a N(108,20) model to approximate the distribution of weekly complaints. Question 3. Tim Cook decrees that the next time the number of complaints received in one week exceeds the value of the third quartile Q3, he will register a strong complaint with the ipod assembly plant in China. What is the value of Q3 (Note: as in question 2, use a N(108,20) model to approximate the distribution of weekly complaints). Use 1 decimal place in your answer.

Answer 1

Question 2)

As the population is normally distributed, we can use standard normal table z to estimate the probability

N(108, 20)

Means

Mean = 108

S.d = 20

We need to find

P(84<X<111) = P(X<111) - P(X<84)

First we need to estimate the z score

Z = (x-mean)/s.d

P(X<111)

Z = (111-108)/20

Z = 0.15

From z table P(Z<0.15) = 0.5596

P(X<84)

Z = (84-108)/20

= -1.2

From z table, P(z<-1.2) = 0.1151

Therefore, the required probability is = 0.5596 - 0.1151

= 0.4445

Question 3)

Q3 is a certain point below which 75% of the data lies

From z table, P(Z<0.67) = 0.75

Therefore, Z = 0.67

0.67 = (x-mean)/s.d

X = 0.67*20 + 108

X = 121.4