Question

Help with 2 statistics problems:
A seller of a machine claims that the machine cuts lumber to a mean length of 96in with a standard deviation of 0.35in. Assume the lengths are normally distributed and assume that the seller is correct.
a. What is the probability that a single randomly selected peice of lumber would have a length greater than 96.15inches?

b. What is the probability that the average length of 49 randomly selected pieces of lumber would be greater than 96.15in ?

I don't need help solving the problems, just setting them up.

Answer 1

For A I have started by: P(Y>96.15) = P(Z> [96.15-96]/0.35) = 1-0.6664 = 0.3336
I'm not sure if I have set this up correctly.

Answer 2

you are trying to find a zscore for 96.15 from the mean ... right?

Answer 3

the z score should be z= 0.42857

Answer 4

im wondering if the minus is due to a left tailed table ...

Answer 5

yes because her table is left tailed

Answer 6

P(Z>.42857) = 1 - P( Z < .42857)

Answer 7

I use a calculator and got the exact answer normalcdf ( 96.15,1E99, 96, .35) =0.3341176, which is 0.334 to three decimals

Answer 8

same here :)

Answer 9

Is this for part A? What is confusing to me is that one is asking for a single sample and the other is asking for the entire mean. . . I'm not sure what the different formulas are

Answer 10

divide the sd by sqrt(n)

Answer 11

\[z=\frac{x-\bar x}{\sigma/\sqrt{n}}\]

Answer 12

in other words, your z score changes by a multiple of \(\sqrt{49}=7\)

Answer 13

AH! I see now, thank you!

Answer 14

youre welcome :)

Answer 15

the new z score for the 49 lumber pieces is z = 3

Answer 16

P(Z>3) = .00134996