Question

The average American man consumes 9.5 grams of sodium each day. Suppose that the sodium consumption of American men is normally distributed with a standard deviation of 0.8 grams. Suppose an American man is randomly chosen. Let X = the amount of sodium consumed. Round all numeric answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N( , )

b. Find the probability that this American man consumes between 8.6 and 9.5 grams of sodium per day.

c. The middle 20% of American men consume between what two weights of sodium? Low: High:

Answer 1

= 9.5 , = 0.8

a)

here x follows normal distribution

X ~ N( , )

X ~ N( 9.5, 0.64 )

b)

We want to find P( 8.6 < x < 9.5)

P( 8.6 < x < 9.5) = P(-1.125 < Z < 0 )

P( 8.6 < x < 9.5) = P(z < 0) - P(z < -1.125)

P( 8.6 < x < 9.5) = 0.5−0.1303

P( 8.6 < x < 9.5) = 0.3697

Probability = 0.3697

c)

we want to find the two points such that the middle 30% of American men consuming sodium

P(-z < Z < z) = 0.20

P(-z < Z < z) = P(Z< z) - P(Z < -z)

P(-z < Z < z) = P(Z< z) - ( 1 - P(Z < z))

P(-z < Z < z) = 2 * P(Z< z)

2 * P(Z< z) -1 =0.20

2 * P(Z< z) = 1 + 0.20

P(Z< z) = 1.20 / 2

P(Z< z) = 0.60

now find the

value of z for P (Z < z) = 0.60

we get z = 0.25

X1 = 9.5 - ( 0.25 * 0.8)

X1 = 9.3

X2 = 9.5 + ( 0.25 * 0.8)

X2 = 9.7

Low: 9.3

High: 9.7