Thread: [SOLVED] Sample Mean and confidence interval

Originally Posted by Sashikala

Please help me on the following problem.
The error made when a certain instrument is used to measure the body length of a butterfly of a particular species is known to be normally distributed with mean 0 and standard deviation 1mm.
(a) Calculate, to three decimal places, the probability that the error made when the instrument is used once is numerically less than 0.4 mm.
(b)Given that the body length of a butterfly is measured nine times with the instrument, calculate, to three decimal places, the probability that the mean of the nine readings will be within 0.5mm of the true length.
(c)Given that the mean of the nine readings was 22.53 mm, determine a 98 % confidence interval for the true body length of the butterfly.

Given, mean=0 and SD=1, it's a standard normal. You can refer to the statistic table directly.
a) The error made is numerically less than 0.4 mm
Therefore, P(|z| less than 0.4)

b) Given that the body length of a butterfly is measured nine times with the instrument, calculate, to three decimal places, the probability that the mean of the nine readings will be within 0.5mm of the true length.

1st: P(|z| less than 0.5) ... this is the probability for one time.
2nd: Let, P(|z| less than 0.5)= . Therefore, the probability is

c) Sorry.... I don't know what is confidence interval...

Thanks very much.

I could not get the answer for (b).
The answer for (b) should be 0.866.
Can anyone help me on (b)?

I got the answer for (b) as well.

Originally Posted by Sashikala

I got the answer for (b) as well.

Is my solution wrong?
Can you tell me what is confidence interval?

Hi SengNee,

Your solution for part (a) is correct. With boost of (a) I worked the
rest.
(b) P( -0.5/(1/3) < z <+0.5/(1/3) )
= 2*0.9332 - 1
= 0.8664

(c) 98% confidence interval is
[(X - 2.3263*sigma/RT(n)) , (X + 2.3263*sigma/RT(n))]
ie [(22.53-2.3263*1/3), (22.53-2.3263*1/3)]
=[21.75,23.31]