Question

The mean cholesterol levels of women age 45-59 in Ghana, Nigeria, and Seychelles is 5.1 mmol/l and the standard deviation is 1.0 mmol/l. Assume that cholesterol levels are normally distributed and a random sample of 24 women are selected.

It is possible with rounding for a probability to be 0.0000.

a) Identify the individual, variable, type of variable and the random variable X in the context of this problem. The individual is

The variable information collected from each individual is Select an answer This variable is

The random variable X is as follows:

b) List the givens with the correct symbols.

? = 5.1 mmol/l

? = 1 mmol/l

? = 24

c) Identify the random variable ¯¯¯ X in the context of this problem.

d) Find the mean of the sampling distribution of the sample mean. Put the numeric value in the first box and the correct units in the second box.

e) Find the standard deviation of the sampling distribution of the sample mean. Put the numeric value rounded to two decimal places in the first box and the correct units in the second box.

f) What is the shape of the sampling distribution of the sample mean?

Why? Check all that apply:

n is less than 30 n is at least 30

population is not normal

σ is known

population is normal

σ is unknown

g) Find the probability that the sample mean cholesterol level of the 24 randomly selected women age 45-59 in Ghana, Nigeria, and Seychelles is less than 5.3 mmol/l.

Round final answer to 4 decimal places.

DO NOT use the rounded standard deviation from part e in this computation.

Use the EXACT value of the standard deviation with the square root.

h) Find the probability that the sample mean cholesterol level of the 24 randomly selected women age 45-59 in Ghana, Nigeria, and Seychelles is more than 5.64 mmol/l.

Round final answer to 4 decimal places.

DO NOT use the rounded standard deviation from part e in this computation.

Use the EXACT value of the standard deviation with the square root.

i) Is a mean cholesterol level of 5.64 mmol/l unusually high for 24 randomly selected women age 45-59 in Ghana, Nigeria, and Seychelles?

j) If you found a mean cholesterol level for a sample of 24 women age 45-59 in Ghana, Nigeria, and Seychelles as high as 5.64 mmol/l, what might you conclude?

Answer 1

a)The individual variable in the above problem is the CHOLESTEROL LEVEL in women of Ghana, Nigeria, and Seychelles  (45-59).

the random variable is the MEAN cholesterol level. it is a continuous variable.

b) for the random variable

= 5.1 mmol/l(population mean)

= 1 mmol/l ( standard dev. of population)

n = 24 (sample size)

c) the question is not complete

d) The mean of the sampling distribution is equal to the mean of the population (μ)= 5.1 mmol/l

e) The standard deviation of the sample mean is the standard deviation of the population divided by the square root of the sample size, that is,

= 1 mmol/l /4.89 = 0.20 mmol/l

f) the question is not complete. it is missing (probably )some figure.

g)

the respective z score with = 5.3

z= 0.9798

Now, we use z-tables to find the probability of less than 0.9798

P(Z<0.9798)=0.16359

h) HINT: similar to part g