Question

A prototype automotive tire has a design life of 38,500 miles with a standard deviation of 2,500 miles. The manufacturer tests 60 such tires. On the assumption that the actual population mean is 38,500 miles and the actual population standard deviation is 2,500 miles, find the probability that the sample mean will be less than 36,000 miles. Assume that the distribution of lifetimes of such tires is normal.

(a) Let X = number of miles on a single tire. Write the question above in terms of this variable X.

(b) Using the software tool above, find the probability stated on part (a)

(c) Using the software tool above, graph the probability of stated on part (b)

2. An automobile battery manufacturer claims that its midgrade battery has a mean life of 50 months with a standard deviation of 6 months. Suppose the distribution of battery lives of this particular brand is approximately normal. On the assumption that the claims are true, find the probability that a randomly selected battery of this type will last less than 48 months. (Use the software link for every question)

(a) Let X = number of months a battery will last. Write the question above in terms of this variable X

(b) Find the probability that a single battery of this type will last less than 48 months.

(c) Find the probability that the mean of a random sample of 36 batteries will be less than 48 months.

(d) Why do you think the values from part (b) and part (c) are different? Explain.

Answer 1

Solution1a:

(a) Let X = number of miles on a single tire. Write the question above in terms of this variable X.

X follows normal ditribution with mean =xbar=38,500 and sd=sigma/sqrt(n)=2500 /sqrt(1)=2500

Xbar~N(38500,2500

Solution-b:

P(Xbar<36000)

Rcode:

library(tigerstats)
pnormGC(bound=36000 ,region="below",mean=38500 ,sd=2500,graph=TRUE)

0.1586553

P(Xbar<36000)=0.1586553

Solution-c:

Rcode:

pnormGC(bound=36000 ,region="below",mean=38500 ,sd=2500,graph=TRUE)

Solution-2a:

X is random variable

distribution of battery lives of this particular brand is approximately normal with mean 50 and sd=6

X~N(50,6)

SOlution-2b:

X~N(50,6/sqrt(1))

P(Xbar<48)

Rcode:

pnormGC(bound=48,region="below",mean=50,sd=6,graph=TRUE)

=0.3694413

Solution-c:

Xbar~N(50,6/sqrt(36)

Xbar~N(50,1)

P(Xbar<48)

pnormGC(bound=48,region="below",mean=50,sd=1,graph=TRUE)

=0.02275013

(d) Why do you think the values from part (b) and part (c) are different? Explain.

when one sample used P(Xbar<48)=0.3694

when 36 samples used P(Xbar<48)=0.0228

As sample size increases,standard deviation decreases,probability decreases