I don’t understand how to calculate a and get the standard deviation for b, the mean for c and e, and don’t get f

Question

The distribution of actual weights of wedges of cheddar cheese produced at a dairy is normal with a mean of 9.1 ounces and a standard deviation of 0.2 ounces. (Round to 2 decimal places for all z-values and round all other answers to 4 decimal places, if needed.)

(a) The probability that a randomly chosen wedge of cheddar cheese is greater than 9.07 is _____.

(b) If a sample of 16 is randomly chosen, then the distribution of the sample mean weight is _____(b2) with a mean of _____ and a standard deviation of _____. (Options for (b2): approximately normal or not normal or left/right skewed)

(c) The probability that the sample mean weight of this sample of 16 is less than 9.07 is ______.

(d) The probability that the sample mean weight of this sample of 16 is greater than 9.07 is ______.

(e) The probability that the sample mean weight of this sample of 16 is between 9.07 and 9.15 is ______.

(f) There is only a 5% chance that the average weight of a sample of these 16 cheese wedges will be below _______.

Answer 1

A normal distribution is described by its mean (mu) and its standard deviation (sigma) and a complex mathematical function. We get its values for any x from tables of the normal distribution as a function of

z = (x - mu)/sigma

For this problem,

mu = 9.1

sigma = 0.2

a. x = 9.07

z = (9.07-9.10)/0.2 = -0.15

this is just under the 50%tile, where z = 0.00

and from the normal distribution tables, N(z=-0.15) = 0.4404 or about 44%

b. samples from a normal distribution produce means that are also distributed approximately normal, with

standard errors = SE = sigma/sqrt(n),

where n = number of samples used to get the mean

and here, n=25, sqrt(n)=5, so SE = 0.2/5 = 0.04

when dealing with sample means, we denote a t statistic rather than the z statistic,

t = (sample mean value - mu) / SE

and use the normal distribution tables for N(t), just as we did for N(z).

c,d,e: calculate the t values and use the normal distribution tables

"an exercise left to the student," the phrase is.