Question

# The average time for a particular LED TV needs ﬁrst service is 415 days with standard...

The average time for a particular LED TV needs ﬁrst service is 415 days with standard deviation of 30 days. The manufacturer gives one year (365 days) onsite service warranty for newly bought TVs. Assume that the times for ﬁrst service call are approximately normally distributed. Let X denote the time until the ﬁrst service for an LED TV.

a. The distribution of X is (i) Standard normal with mean 0 and variance 1. (ii) Normal with mean 415 days and standard deviation 30 days. (iii) Normal with mean 365 days and standard deviation 30 days. (iv) none of the above

b. The probability that a TV requires the ﬁrst service within the warranty period is 126 Chapter 6. Test and Conﬁdence Interval for a Proportion (i) 0.9522 (ii) .0478 (iii) 0.0228 (iv) none of the preceding

c. Determine the warranty period so that the manufacturer is expected to service about 5% of the TVs within the warranty period. (i) 365.7 days; about a year (ii) 400 days (iii) 395 days (iv) 300 days

d. Suppose the warranty time is determined as in part c, and a batch of 5,000 TVs were sold. Let X denote the number of TVs need service within the warranty period. The distribution of X is (i) binomial with n = 5000 and p = 0.95 (ii) binomail with n = 5000 and p = 0.05 (iii) Poisson with mean 4750. (iv) none of the above

e. Suppose the warranty time is determined as in part c, and a batch of 5,000 TVs were sold. Then the probability that 300 or more TVs will require service within the warranty period is (i) 0.0009 (ii) 0.0090 (iii) 0.0011 (iv) none of the preceding

A)

X : Time until the first service call

X ~ Normal(415,302)

Option (ii) is correct

B)

Since μ=415 and σ=30 we have:

P ( X<365 ) = P ((Xμ)/σ<(365−415)/30)

Since (xμ/σ) = Z and (365−415)/30=−1.667 we have:

P (X<365)=P (Z<−1.667) = 0.0478

Option (ii) is correct

(C)

Let the warranty period be R

We have to find P ( X<R) = 0.05

P ((Xμ)/σ<(R−415)/30) = 0.05

P (Z<(R−415)/30) = 0.05

(R−415)/30 = -1.6449

R = -1.6449 * 30 + 415 = 365.7 days

Option (i) is correct

D)

X : Number of TVs need service within the warranty period.

X ~ Binomial(5000,0.05)

Option (ii) is correct

E)

We have to find probability P(X>=300)

= 1 - P(X<300) = 1- 0.99912 = 0.000878 = 0.0009

Option (i) is correct

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