Question

.The lifetimes of a certain brand of LED light bulbs are normally distributed with a mean of 53,400 hours and a standard deviation of 2500 hours.A.If the company making these light bulbs claimed that they would last at least 50,000 hours. What proportion of light bulbs would meet the claim and last at least 50,000 hours? (12)B.The company’s marketing director wants the claimed figure to be where 98% of these new light bulbs to last longer than the amount claimed by the company’s advertising? What number should be claimed so that 98% of the company’s new LEDlight bulbs meet the claim? (12)C.If a random sample of 100 of these light bulbs is selected, what is the probability that the sample mean lifetime, ?̅, will be at least 53,000 hours? please show work

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	53400
std deviation   =σ=	2500.000

A)

proportion of light bulbs would meet the claim and last at least 50,000 hours:

probability =P(X>50000)=P(Z>(50000-53400)/2500)=P(Z>-1.36)=1-P(Z<-1.36)=1-0.0869=0.9131

B)

for top 98 % or 2th percentile critical value of z=		-2.05
therefore corresponding value=mean+z*std deviation=53400-2.05*2500 =			48275.00 hours

C)

sample size       =n=	100
std error=σx̅=σ/√n=	250.0000

  probability that the sample mean lifetime, ?̅, will be at least 53,000 hours:

probability =P(X>50000)=P(Z>(50000-53400)/250)=P(Z>-13.6)=1-P(Z<-13.6)=1-0.0000=1.0000