he indicated probability. 19) A bank\'s loan officer rates applicants for credit

Question

he indicated probability. 19) A bank's loan officer rates applicants for credit. The ratings are normally distribu & A ted with a mean o f 19) 200 and a standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 200 and 275. 0.4332 B) 0.5 C 0.9332 D) 0.0668 4- e the problem. 20) 20) A company manufactures batteries in batches of 15 and there is a 3% rate of defects. Find the standard deviation for the number of defects per batch. A) 0.4 B) 43.7 0.2 D) 0.7 Part II Each question is worth 10 points. All work must be shown. If using a calculator, indicate what formula you are using. 1.) The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days a. One classical use of the normal distribution is inspired by a letter to "Dear Abby" in which a wife claimed to have given birth 308 days after a brief visit from her husband, who was serving in the Navy. Given this information, find the probability of a pregnancy lasting 308 days or longer. What does the result suggest? b. If we stipulate that a baby is premature if the length of pregnancy is in the lowest 496, find the length that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care = 268

Explanation / Answer

Question 19

Ratings are normally distributed.

We are given

Mean = 200

SD = 50

We have to find P(200<X<275)

P(200<X<275) = P(X<275) – P(X<200)

Z = (X – Mean) / SD

For P(X<275)

Z = (275 – 200) / 50 = 75/50 = 1.5

P(Z<1.5) = P(X<275) = 0.933193

For P(X<200)

Z = (200 – 200) / 50 = 0

P(Z<0) = P(X<200) = 0.5

P(200<X<275) = P(X<275) – P(X<200)

P(200<X<275) = 0.933193 – 0.5 = 0.433193

Required probability = 0.4332

Question 20

We are given n = 15, p = 0.03, q = 1 – p = 1 – 0.03 = 0.97

Here, we have to use binomial distribution.

SD = sqrt(npq) = sqrt(15*0.03*0.97) = 0.660681466

Required answer = 0.7

Part II

Question 1.a.

We are given lengths of pregnancies are normally distributed.

Mean = 268

SD = 15

We have to find P(X>308)

P(X>308) = 1 – P(X<308)

Z = (X – mean) / SD

Z = (308 – 268) / 15 = 2.666666667

P(Z<2.666666667) = P(X<308) = 0.996169619

P(X>308) = 1 – P(X<308)

P(X>308) = 1 – 0.996169619

P(X>308) = 0.003830381

Required probability = 0.003830381

The result suggests that there is insufficient evidence to conclude the claim of wife.

Question 1.b.

Let length of pregnancies (X) followed normal distribution with mean 268 and SD = 15 days.

X = Mean + Z*SD

For the lower 4%, critical Z score = -1.750686071

X = 268 + (-1.750686071)*15

X = 241.7397089

Required answer = 241.7397089 days

Question 2.a.

We are given n = 2600, p = 0.077, q = 1 – p = 1 – 0.077 = 0.923

Mean = n*p = 2600*0.077 = 200.2

SD = sqrt(n*p*q) = sqrt(2600*0.077*0.923) = 13.59354994

Question 2.b.

We are given n = 2600, p = 0.077, we have to find P(X= 175)

By using continuity correction and normal approximation to binomial distribution, we have to find P(174.5<X<175.5)

P(174.5<X<175.5) = P(X<175.5) – P(X<174.5)

Z = (175.5 – 200.2) / 13.59 = -1.817038236

P(Z< -1.817038236) = 0.034605629

Z = (174.5 – 200.2) / 13.59 = -1.890602537

P(Z< -1.890602537) = 0.029338711

P(174.5<X<175.5) = P(X<175.5) – P(X<174.5)

P(174.5<X<175.5) = 0.034605629 - 0.029338711

P(174.5<X<175.5) = 0.005266919

The above probability is less than 0.05. So this probability is unusual.

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