Suppose the mean daily caloric intake per person in the US is 2300 calories per

Question

Suppose the mean daily caloric intake per person in the US is 2300 calories per day, with a standard deviation of 215 calories. A researcher is planning to take a random sample of 40 people. 1) What is the probability that the sample mean caloric intake per day will be less than 2250 calories?
2) There is a 20% chance the researcher’s sample mean will be below what value?
3) How many total calories would you need to be 90% sure you could provide the caloric intake of 40 randomly chosen people for one day?
4) In answering problems 1, 2, and 3, how did we know that the sample mean is normally distributed?

Explanation / Answer

mean = 2300
std.dev. = 215
n = 40

1)
P(X < 2250) = P(z < (2250 - 2300)/(215/sqrt(40)) = P(z < -1.4708) = 0.0707

2)
z-value = -0.8416
xbar = mean + z*sigma/sqrt(n)
xbar = 2300 -0.8416 * 215/sqrt(40)
xbar = 2271.39

3)
z-value = 1.28
xbar = mean + z*sigma/sqrt(n)
xbar = 2300 + 1.28 * 215/sqrt(40)
xbar = 2343.5129

4)
As the sample size is greater than 30, we can apply central limit theorem. Hence the principles of normal distributions are used.

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