# HELP [Confused] Case: Fuji Film Introduces Aps

• yanip

#### yanip

HELP! [Confused] Case: Fuji Film Introduces Aps

1. As stated, by 1998 APS cameras owned 20% of the point and shoot camera market. Now it is the year 2003 and the market share might be nearer to 40%. Suppose 30 cusomers from the point and shoot camera market are randomly selected. If the market share is really .40 , what is the expected number of point and shoot camera customers who purchase an APS camera? What is the probability that six or fewer purchases and APS camera? Suppose you actually got six or fewer APS customers in the sample of 30. Based on the probability just calculated, is this enogh evidence to convince you that the market share is 40% Why or why not?

2. Suppose customer complaints on the 24 millimeter film are poison distributed at an avegare rate of 2.4 complaints/100,000 rols sold. Suppose further that Fuji is having trouble with shipments being late and one batch of 100,000 rolls yields seven complaints from customers. Assuming that it is unacceptable to management for the average rate of complaints to increase, is this enough evidence to convince management that the average rate of complaints has increased, or can it be written off as a random occurence that happens quite frequently? Produce the Poisson distribution for this question and discuss its inplication for this problem.

3. One study of 52 product launches found that those undertaken with revenue growth as tha main objective are more likely to fail than those undertaken to increase customer satisfaction or to create a new market such as the APS system. Suppose of the 52 products launched, 34 were launched with revenue growth as the main objective and the rest were launched to increase cusotmer satisfaction or to create a new market. Now suppose only 10 of these products were successful ( the rest failed) and seven were products that were launched to increase customer satisfaction or to create a new market. What is the probability of this result occurring by chance? What does this probability tell you about the basic premise regarding the importance of the main objective?

Welcome to the PF, yanip. You are required to show your own work on homework problems before we can help you. What are the relevant equations and concepts involved in calculating these probabilities? How would you go about applying them?

I use minitab but tried to solve using the formula.
For Q1:
I got these answers because P(x)= nCx * p^x * q^ n-x
n=30, x<=6, p=.4
Meaning
30 nCr 6 *(.40)^6 * (.60)^24= .0116
30 nCr 5 *(.40)^5 * (.60)^25= .0041
30 nCr 4 *(.40)^4 * (.60)^26= .0012
30 nCr 3 *(.40)^3 * (.60)^27= .0000
30 nCr 2 *(.40)^2 * (.60)^28= .0000
30 nCr 1 *(.40)^1 * (.60)^29= .0000
30 nCr 0 *(.40)^0 * (.60)^30= .0000

I don't know how to figure out the graph through mini tab and to solve Q3

Thank you.

solution of 1st que

n=30
p=.4
1-p=.6
expected=12
p(atleast six)=p(<=6)=.0172

I use minitab but tried to solve using the formula.

For Q1:
I got these answers because P(x)= nCx * p^x * q^ n-x
n=30, x<=6, p=.4
Meaning
30 nCr 6 *(.40)^6 * (.60)^24= .0116
30 nCr 5 *(.40)^5 * (.60)^25= .0041
30 nCr 4 *(.40)^4 * (.60)^26= .0012
30 nCr 3 *(.40)^3 * (.60)^27= .0000
30 nCr 2 *(.40)^2 * (.60)^28= .0000
30 nCr 1 *(.40)^1 * (.60)^29= .0000
30 nCr 0 *(.40)^0 * (.60)^30= .0000

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