Question

In: Statistics and Probability

The quality control manager of​ Marilyn's Cookies is inspecting a batch of​ chocolate-chip cookies that has...

The quality control manager of​ Marilyn's Cookies is inspecting a batch of​ chocolate-chip cookies that has just been baked. If the production process is in​ control, the mean number of chip parts per cookie is 6.7. Complete parts​ (a) through​ (d).

A. What is the probability that in any particular cookie being inspected fewer than five chip parts will be found?

B. What is the probability that in any particular cookie being inspected exactly five chip parts will be found?

C. What is the probability that in any particular cookie being inspected five or more chip parts will be found?

D. What is the probability that in any particular cookie being inspected either four or five chip parts will be found?

Solutions

Expert Solution

Given,
Mean number of chip parts per cookie: = 6.7

X : Number of chip parts in a cookie

X follows Poisson distribution with = 6.7 and the probability distribution of X

A.

Probability that in any particular cookie being inspected fewer than five chip parts will be found; P(X<5)

P(X<5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)

P(X<5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 0.0012+0.0083+0.0276+0.0617+0.1034=0.2022

Probability that in any particular cookie being inspected fewer than five chip parts will be found = 0.2022

B.

Probability that in any particular cookie being inspected exactly five chip parts will be found = P(X=5)

Probability that in any particular cookie being inspected exactly five chip parts will be found = 0.1385

C.

Probability that in any particular cookie being inspected five or more chip parts will be found = P(X5)

P(X5) = 1 -P(X<5)

From A . P(X<5) = 0.2022

P(X5) = 1 -P(X<5) = 1- 0.2022 = 0.7978

Probability that in any particular cookie being inspected five or more chip parts will be found = 0.7978

D.

Probability that in any particular cookie being inspected either four or five chip parts will be found = P(X= 4 or 5)

P(X= 4 or 5) = P(X=4) + P(X=5)

P(X=4) + P(X=5) = 0.1034+0.1385=0.2419

P(X= 4 or 5) = P(X=4) + P(X=5) = 0.2419

Probability that in any particular cookie being inspected either four or five chip parts will be found = 0.2419


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