In 2016, the CDC estimated the mean weight of U.S. women over the age of 20 years old was 168.5 pounds with a standard deviation of 68 pounds.

1. What is the expected mean for a sample of 150 women?
2. What is the standard deviation of the mean for a sample of 150 women?
3. What is the probability of 150 women having a sample mean below 160 pounds?
4. What is the probability of 150 women having a sample mean above 175 pounds?
5. What is the probability of 200 women having a sample mean below 160 pounds? Note the change in sample size.
6. What is the probability of 200 women having a sample mean above 175 pounds?

Respuesta :

Step-by-step explanation:

1.

the expected sample mean is always the general mean : 168.5 pounds.

2.

the SD of a sample is the general SD / sqrt(sample size).

in our case

the sample SD = 68/sqrt(150) = 5.55217675...

3.

if we are looking for only the probability that any single woman is below 160 pounds, we would use the normal z calculation :

z = (x - mean)/SD = (160 - 168.5)/68 = -8.5/68

but we have here the question about the probability of the mean value of a whole sample of 150 women.

so, we need to adapt the z-calculation by the principle of 2) for the SD of a sample :

z = (x - mean)/(SD × sqrt(sample size)) =

= (160 - 168.5)/(68 × sqrt(150)) = -8.5/(68×sqrt(150)) =

= -0.010206207 ≈ -0.01

that gives us in the z-table the p-value 0.49601

this 0.49601 is the probability that a sample of 150 women has a mean value of below 160 pounds.

4.

similar to 3.

the z value we are looking for

z = (175 - 168.5)/(68 × sqrt(150)) = 6.5/(68 × sqrt(150)) =

= 0.007804747... ≈ 0.01

that gives us the p-value 0.50399.

that would be the probability of a sample mean of 175 or below.

to get above 175 we need to get the other side of the bell-curve :

1 - 0.50399 = 0.49601

so, this case has about the same probability as 3.

5.

as 3), just with the sqrt(200) instead of the sqrt(150).

z = -8.5/(68 × sqrt(200)) = -0.008838835... ≈ 0.01

so, the probability is still about the same as in 3) :

0.49601

6.

as 4) just with sqrt(200).

z = 6.5/(68 × sqrt(200)) = 0.006759109... ≈ 0.01

so, the probability is still about the same as for 4) :

0.49601

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