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.

Required:
What is the probability that between 77 and 120 complaints are received in one week?

Answer 1

0.6856

Explanation:

`\text{The missing part of the question states that we should Note: that N(108,20) model to } \\ \\ \text{ } \text{approximate the distribution of weekly complaints).]}`

Now; assuming X = no of complaints received in a week

Required:

To find P(77 < X < 120)

Using a Gaussian Normal Distribution (
`\mu =`108,
`\sigma` = 20)

Using Z scores:

`Z = (77-108)/(20) \\\\ Z = -(35)/(20) \\ \\ Z -1.75`

As a result X = 77 for N(108,20) is approximately equal to to Z = -1.75 for N(0,1)

SO;

`Z = (120-108)/(20) \\ \\ Z = (12)/(20)\\ \\ Z = 0.6`

Here; X = 77 for a N(108,20) is same to Z = 0.6 for N(0,1)

Now, to determine:

P(-1.75 < Z < 0.6) = P(Z < 0.6) - P( Z < - 1.75)

From the standard normal Z-table:

P(-1.75 < Z < 0.6) = 0.7257 - 0.0401

P(-1.75 < Z < 0.6) = 0.6856