Question

Question 2 (5 points) Suppose that a sample of size 58 is drawn from a population with mean 55 and standard deviation 38 . Find the value of o,, the standard deviation of the distribution of sample means. Write only a number as your answer. Round to two decimal places (for example: 8.21). Your Answer Answer Question 3 (5 points) The weights of bowling balls are normally distributed with mean 11.5 pounds and standard deviation 2.7 pounds. A sample of 36 bowling balls is selected. What is the probability that the average weight of the sample is less than 11.07 pounds? Write only a number as your answer. Round to 4 decimal places (for example 0.0048). Do not write as a percentage Your Answer

Answer 1

Answer

(2) We know that the relationship between the population standard deviation and sample standard deviation $sigma_x$ is given as

$sigma_x$ =

n is sample size

we have the following values   =38 and n = 58

setting the values, we get

38/v/58 38/7.616 4.99

So, required sample standard deviation is 4.99

(3) It is given that

mean

μ = 11.5

standard deviation

sample size n = 36

we have to find the probability that the average weight of a sample is less than 11.07

so,

11.07

Using the formula

setting the given values, we get

P(< 11.07) = P(z < (11.07-11.5)/(2.7/V36)) = P(z < (-0.43)// (0.45))

this gives us

P(< 11.07) P(z <-0.9556)

using the identity

we can write it as

P(z <-0.9556) = 1-P(z < 0.9556)

Using the z distribution table for the 0.9556 value, we get

P(z <-0.9556) = 1-P(z < 0.9556)-1-0.8315= 0.1685

So, the required p value is 0.1685