Question

Let X denotes lifetime of lamp brand with average of 25 months and the variance 400. If we assume that the pattern of lamp lifetime follows normal distribution, calculate the below items:

a) What is the probability that a lamp lasted less than 45 months?

b) What is the probability that a student scored at least 45 months?

c) What is the probability that a lamp worked within 1 standard deviation of the mean?

d) If a lamp lifetime is above the 80th percentile, the company get excellent rate in review, what value it must achieve to get an excellent rate?

e) If a school purchase 30 of these lamps, what is the probability that 6 out of them will receive excellent rate?

Answer 1

here standard deviation =sqrt(400)=20

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	25
std deviation   =σ=	20.0000

probability that a lamp lasted less than 45 months:

probability =

P(X<45)

=

P(Z<1)=

0.8413

0.1587

b)

probability that a student scored at least 45 months =P(X>45)=1-P(X<45)=1-0.1587 =0.8413

c)

probability that a lamp worked within 1 standard deviation of the mean =P(-1<Z<1)=0.8413-0.1587 =0.6826

d)

for 80th percentile critical value of z=		0.84
therefore corresponding value=mean+z*std deviation=			41.80

e)P(6 will receive excellent rating)=³⁰C₆(0.2)⁶(0.8)²⁴ =0.1795