A company that produces DVD drives has a 6% defective rate. Let X represent the

Question

A company that produces DVD drives has a 6% defective rate. Let X represent the number of defectives in a random sample of 59 of their drives.

a. What is the probability the sample will contain exactly 8 defective drives? Give your answer to four decimal places.

b. What is the probability the sample will contain more than 8 defective drives? Give your answer to four decimal places.

c. What is the probability the sample will contain less than 8 defective drives? Give your answer to four decimal places.

d. What is the expected number of defective drives in the sample? Give your answer to two decimal places.  

e. What is the variance of the number of defective drives in the sample? Give your answer to four decimal places.

f. What is the standard deviation of the number of defective drives in the sample? Give your answer to four decimal places.

g. Each defective drive costs the company 12 dollars. What is the expected cost to the company for the defective drives in the sample? Give your answer to two decimal places.  

h. Each defective drive costs the company 12 dollars. What is the standard deviation of the cost to the company for the defective drives in the sample? Give your answer to four decimal places.

Explanation / Answer

Solution:-

A company that produces DVD drives has a 6% defective rate. Let X represent the number of defectives in a random sample of 59 of their drives.

p = 6/100 = 0.06

n = 59

a) The probability the sample will contain exactly 8 defective drives is 0.0159.

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

P(x = 8) = 0.0159.

b) The probability the sample will contain more than 8 defective drives is 0.0083.

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

P(x > 8) = 0.0083.

c) The probability the sample will contain less than 8 defective drives is 0.9759.

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

P(x < 8) = 0.9759

d) The expected number of defective drives in the sample is 3.54.

E(x) = n × p

E(x) = 59 × 0.06

E(X) = 3.54

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