求救統計學

求救統計學

[匿名]

2008-06-24 4:15 am

Q 1

In a production process, the standard length of a test tube is 4.2 inches.
From past measurements, the length of the test tube is a random variable
having a normal distribution with a mean of 4.2 inches and a standard
deviation of 0.1 inch.

a) What is the probability that a randomly selected test tube will have a
length longer than 4.35 inches?

b) A random sample of 20 test tubes was taken.

i. What are the mean and the standard deviation of the corresponding sample mean?

ii. What is the probability that the sample mean length of the 20 test tubes will have a length between 4.15 and 4.25 inches?

iii. Find the length that would be exceeded by 85% of the sample
means.

Q 2

The production manager needs to estimate the amount of time required
for a production line to assemble a garden mower. A random sample of
the amount of time in minutes spent in assembling 18 mowers is given as
follows:

17 18 21 16 22 20 24 22 31 18 20 24 25 20 24 26 28 25

a) In constructing a confidence interval for the true average time required to assemble a mower, elaborate which distribution has to be used. What assumption(s) must be made initially?

b) Construct a 95% confidence interval estimate for the true average time required to assemble a mower. Keep the corresponding answers to one decimal place.

回答 (1)

凡人細語

2008-06-25 2:07 am

✔ 最佳答案

q1)

a. Let X be length of test tube which is in N(4.2, 0.01)

Pr(X>4.35)= P((X-4.2/.1> (4.35-4.2)/.1)

( convert all Normal distribtuion to Standard Normal Distribution)

we get Pr(z > 1.5)

= 1- Pr(z<=1.5)

=1-.93319( use table or excel function)

=.06681

b(i)

you should note from the sampling distribution:

if X is N(u, (sigma)^2)

then barX is N( u, (sigma)^2/n) where n is the sample number, and Bar X is average of sample value

so the mean is just 4.2 ( average value of the sample mean is the population mean)

and the variance is (sigma)^2/n= 0.01/20

standard devation is just sqrt( 0.01/20)= 0.0224
ii
now we know bar(x) is N(4.2, .01/20)

now Pr( 4.15<bar(x)<4.25)=Pr( (4.15-4.2)/.0224 <z < (4.25-4.2)/.0224)
=Pr(- 2.232142857< z<2.232142857)
=Pr( z<2.232142857) - Pr(z <- 2.232142857)

- =0.98713- (1- 0.98713) ( by symmetry)
=0.97426

iii.

let Y be the length required:

we are interested in finding y such that

Pr( Y<bar(x))= 0.85

so Y is within 1.04 standard devation from the mean (4.2)

i.e Y= 1.04(0.0224) 4.2 = 4.203296

Q2)

first asusme it is normally distributed, asumpiton: population varaicne is known and n is large enough for normal distribtuion justify( actually since n<23, so we should use student-t distribtuion instead)

b) if normally distributed,

CI for average time= bar(x) /- 1.96*((sigma(x)

where you find bar(X) and sigma(X) yourself, it is just very simple

If t distributed,

CI for avarge time= bar(x) /- t(0.025)(sigmax)

where t(.025) is 2.101 with degree of freedom v=18

good luck! ^^

參考: Statistical Tables for Examinations

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