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 0.9 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 10.4 and 11.2 grams of sodium per day.

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

Answer 1

Solution :

Given that ,

mean = $\mu$ = 9.5

standard deviation = $\sigma$ = 0.9

a) The distribution of x is normal

b) P(10.4 < x < 11.2) = P[(10.4 - 9.5)/ 0.9 ) < (x - $\mu$ ) /$\sigma$  < (11.2 - 9.5) / 0.9) ]

= P(1.00 < z < 1.89)

= P(z < 1.89) - P(z < 1.00)

Using z table,

= 0.9706 - 0.8413

= 0.1293

c) Using standard normal table,

P( -z < Z < z) = 20%

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

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

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

= P(Z < z) = 1.20 / 2

= P(Z < z) = 0.60

= P(Z < 0.25) = 0.60

= z  ± 0.25

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -0.25 * 0.9 + 9.5

x = 9.3

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 0.25 * 0.9 + 9.5

x = 9.7

The middle 20% are from 9.3 to 9.7

low = 9.3

high = 9.7