Suppose we took random samples of 30 watches from the population. Assume the pop

Question

Suppose we took random samples of 30 watches from the population. Assume the population standard deviation is eight months ( 0.67 years) and a mean of 4 years

a. What is the mean, standard deviation and shape of the sampling distribution (the sample means)?

b. What is the probability that the sample mean of the 30 watches will be between 3.8 years and 4.2 years?

c. What is the probability that the sample mean will be longer than 4.3 years? c.

d. Find a value, C, such that P(x>= C) = 0.15. In other words, what is the 85th percentile of the means?

Only part (a) to (d) please alone Please need it ASAP. All questions to be answered well in full

everthing is given please . this a normally distributed question

Back to our Timely Brand Watch Company above... suppose we took random samples of 30 watches from the population. Assume the population standard deviation is eight months.) a. What is the mean, standard deviation and shape of the sampling distribution (the sample means)? b. What is the probability that the sample mean of the 30 watches will be between 3.8 years and 4.2 years? What is the probability that the sample mean will be longer than 4.3 years? c. d. Find a value, C, such that P(x2 C) = .15. In other words, what is the 85th percentile of the means? If we had a population that was normally distributed and had a mean of 50 and standard deviation of 5.... a. If we sampled a lot of items, say 100 items, from this population, what shoul we expect the largest and the smallest values to be in that sample? (Use the

Explanation / Answer

SolutoonA:

according to central limit theorem

sample mean=popuation mean=4 years

sample standard deviation/=population std devsqrt(sample size)

=0.67/sqrt(30)

=0.1223247

Solutionb:

P(3.8<X<4.2)

p(3.8-4/0.1223247<Z<4.2-4/0.1223247)

P(-1.634993<Z<1.634993)

p(z<1.634993)-p(z<-1.634993)

pnorm(1.634993)-pnorm(-1.634993)

=0.8979495

ANSWER:0.8979495

Solutionc:

P(X bar>4.3)

P(4.3-4/0.1223247)

=2.452489

pnorm(-2.452489)

=0.007093586

ANSWER:0.007093586

Solutiond:

z for 85 percentile is

1.036

pnorm(-1.036)

=0.1501011

P(Z>-1.036)=0.1501011

C=1.036

X-mean/sd=1.036

x=1.036*0.1501011+4

X= 4.155505

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