A process produces strings of Christmas tree lights that historically have exper

Question

A process produces strings of Christmas tree lights that historically have experienced a detective rate of 4%. A customer has placed an order for 150 strings of lights. Use the normal approximation to the binomial distribution to answer the following: a. Calculate the mean and standard deviation for this distribution. b. What is the probability that less than 2 strings in this order will be detective? c. What is the probability that exactly 7 strings in this order will be detective? d. What is the probability that 3, 4, or 5 strings in this order will be detective? e. What is the probability that 8, 9, or 10 strings in this order will be detective?

Explanation / Answer

probability for the strings to be defective , p = .04 or 4%,

probability for the strings to be not defective, q = 0.96 or 96%

We can approximate the distribution to normal if np and nq >5, where n is no of observations = 150

np = 150*0.04 = 6 and nq = 150*.96 = 144.

Hence we can approximate it to a normal distribution

a. Mean and standard deviation

Mean and standard deviation are given by the binomial distribution as follows

Mean , mu = np = 150*0.04 = 6

standard deviation, sd = npq = 150*0.04*0.96 = 5.76

We use normal distribution to find values for the rest of the questions, we will be doing this by calculating the zcores and getting the probabilities from the normal distribution table.

Also we need to convert the discrete distribution to continuous distribution. This can be arrived as follows

x<2 can be written as x<1.5, x=7 can be written as 6.5<x<7.5 etc

b. Probability of less than two strings to be defective

P(x<2) = P(x<1.5)

z-score = (x-mu)/sd = (1.5-6)/5.76 = -0.781

from normal distribution table , P(z<0.781) = 0.217

Hence the probability that less than two strings are defective = 0.217

c. probability that exactly 7 strings are defective

P(x=7) = P(6.5<x<7.5)

zscore at x= 6.5 = (6.5 - 6)/ 5.76 = 0.087

Similarily zscore at x =7.5,= 1.5/5.76 = 0.260

Hence , P(6.5<x<7.5) = P(z<0.260) - P(z<0.087) = 0.603-0.535 = 0.068

Probability that exactly 7 strings are defective = 0.068

d. Probability that 3,4 or 5 strings are defective

P(x=(3,4,5)) = 1- P(x<2.5)-P(x>5.5)

Since total probability is always 1, P(x=(3,4,5)) can be written as 1- P(all other outcomes)

zscore of 2.5 = 2.5-6 / 5.76 = -0.608

zscore of 5.5 = -0.5/5.76 = -0.087

From normal dist. table, P(z<-0.608) = 0.272

P(z< -0.087) = 0.465

hence P(z>-0.087) = 1-0.465 = 0.535

Hence ,  P(x=(3,4,5)) = 1 - 0.272 - 0.535 = 0.193

Probability that 3,4 or 5 strings are defective = 0.193

e. Probability that 8,9 or 10 strings are defective

P(x=(8,9,10)) = 1- P(x<7.5)-P(x>10.5)

zscores : 7.5 = 1.5/5.76 = 0.260: 10.5 = 4.5/5.76 = 0.781

P(z<0.260) = 0.603

P(z<0.781) = 0.783

P(z>0.781) = 1-0.783 = 0.217

P(x=(8,9,10)) = 1-0.603-0.217 = 0.18

Probability that 8,9 or 10 strings are defective = 0.18

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