Question

calculate PART A, PART B, PART C:

PART A: An electronic product takes an average of 3 hours to move through an assembly line. If the standard deviation of 0.5 hours, what is the probability that an item will take between 1.7 and 3.3 hours to move through the assembly line?

(Round your answer to 3 decimal places.)

PART B: A manufacturer knows that their items have a normally distributed lifespan, with a mean of 2.5 years, and standard deviation of 0.7 years.

If 25 items are picked at random, 1% of the time their mean life will be less than how many years?

Give your answer to one decimal place.

PART C: Delivery times for shipments from a central warehouse are exponentially distributed with a mean of 2.63 days (note that times are measured continuously, not just in number of days). A random sample of 143 shipments are selected and their shipping times are observed.
Approximate the probability that the average shipping time is less than 2.29 days.
Enter your answer as a number accurate to 4 decimal places.

Answer 1

A)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	3
std deviation   =σ=	0.5000

probability =

P(1.7<X<3.3)

=

P(-2.6<Z<0.6)=

0.7257-0.0047=

0.7210

B)

mean μ=	2.5
standard deviation σ=	0.7000
sample size       =n=	25
std error=σx̅=σ/√n=	0.14

for 1th percentile critical value of z=		-2.33
therefore corresponding value=mean+z*std deviation=			2.2

C)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	2.63
std deviation   =σ=	2.6300
sample size       =n=	143
std error=σx̅=σ/√n=	0.2199

probability that the average shipping time is less than 2.29 days :

probability =

P(X<2.29)

=

P(Z<-1.55)=

0.0606