Question

In: Statistics and Probability

In the United States, males between the ages of 40 and 49 eat on average 105.1...

In the United States, males between the ages of 40 and 49 eat on average 105.1 g of fat every day with a standard deviation of 4.5 g ("What we eat," 2012). Assume that the amount of fat a person eats is normally distributed.

(a) State the random variable.

(b) Find the probability that a man age 40-49 in the U.S. eats more than 110 g of fat every day.

(c) Find the probability that a man age 40-49 in the U.S. eats less than 93 g of fat every day.

(d) Find the probability that a man age 40-49 in the U.S. eats between 65 g and 100 g of fat every day.

(e) If you found a man age 40-49 in the U.S. who says he eats less than 65 g of fat every day, would you believe him? Why or why not?

(f) What daily fat level do 95% of all men age 40-49 in the U.S. eat less than?

Solutions

Expert Solution

(a) State the random variable..

X: Daily fat consumption of male in US

b)

The following information has been provided:

We need to compute The corresponding z-value needed to be computed is:

Therefore, we get that

c)

We need to compute . The corresponding z-value needed to be computed:

Therefore,

d)

We need to compute . The corresponding z-values needed to be computed are:

Therefore, we get:

e)

We need to compute . The corresponding z-value needed to be computed:

Therefore,

.no, since the probability is 0

f)

We need to compute x such that . The corresponding z-value is 1.645

x = 105.1 = 7.4025

x = 105.1 + 7.4025

x = 112.5025 g

.


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