Question

The average American man consumes 9.8 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 10.6 and 11.1 grams of sodium per day.

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

Low:

High:

Answer 1

solution

mean = = 9.8

standard deviation = = 0.8

(a) X ~ N(9.8 , 0.64

b)P(10.6 < x < 11.1) = P[(10.6 - 9.8)/0.8 ) < (x - ) /  < (11.1 - 9.8) /0.8 ) ]

= P(1 < z <1.63 )

= P(z < 1.63) - P(z <1 )

Using z table,

= 0.9484 - 0.8413

answer =0.1071

(c) middle 20% of score is

P(-z < Z < z) = 0.20

P(Z < z) - P(Z< -z) = 0.20

2P(Z < z) - 1 = 0.20

2P(Z < z) = 1 + 0.20 = 1.20

P(Z < z) = 1.20/ 2 = 0.60

P(Z < 0.253) = 0.84

middle 68% has two z value -1 and +1

z =- 0.253

z = 0.253

Using z-score formula,

x = z * +

x = -0.253* 0.8 + 9.8

x = 9.5976

Using z-score formula,

x = z * +

x = 0.253* 0.8 + 9.8

x = 10.0024

low = 9.5976

high = 10.0024