2. A population has a normal distribution with a mean of 130 and a standard devi
ID: 2957450 • Letter: 2
Question
2. A population has a normal distribution with a mean of 130 and a standard deviation of 30. Find the probability that a single element selected from the population will have a value between 139.50 and 167.25.U= 130 and s= 30
P(139.50 < x< 167.25 = p ( 139.50 -130/30 < x-u/s< 167.25-130/30)
= P (0.32 <z< 1.24)
= (1.24) – (0.32)
= 0.8925- 0.6255
= 0.2670
The probability that a single element selected from the population will have a value between 139.50 and 167.25 is 0.2670
3. Use the population described in problem 2 above. Now find the probability that the sample mean for a sample of 16 elements selected from the population will be between 139.50 and 167.25.
u=130,s=30, n =16
s/vn= 30/v16= 7.5
p( 139.50 <x< 167.25)
=p( 139.50-130/ 7.5 < x-u/s/vn<167.25-130/ 7.5
= p
=P (1.27<z<4.97)
= (4.97)- (1.27)
= 0.9999- 0.898
= 0.1019
Probability that the sample mean for a sample of 16 elements selected from population will be between 139.50 and 167.25 is 0.1019
4. Explain the reason the answers to 2 and 3 above are different.
Explanation / Answer
In question no. 2, population mean and standard deviation are given hence, formula of z is x-u/s but in case of question no. 3, sample size comes into picture because of which we need to introduce and consider the standard error which provides an unbiased estimate of the standard deviation and formula of standard error is s/vn and hence formula of z is x-u/s/vn
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