Question

The average American man consumes 9.7 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. Find the probability that this American man consumes between 8.8 and 9.8 grams of sodium per day

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

Answer 1

Given that, mean (μ) = 9.7 grams and

standard deviation = 0.9 grams

Let X = the amount of sodium consumed.

Here, X ~ Normal (9.7, 0.9)

A) We want to find, P(8.8 < X < 9.8)

Therefore, the probability that this American man consumes between 8.8 and 9.8 grams of sodium per day is 0.3851

B) We want to find, the values of x1 and x2 such that,

P(x1 < X < x2) = 0.20

First we want to find, the z-score such that, P(-z < Z < z) = 0.20

P(-z < Z < z) = 0.20

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

=> 2 * P(Z < z) = 1.20

=> P(Z < z) = 0.6

Using Excel we get, z-score corresponding probability of 0.6 is, z = 0.2534

Excel Command : = NORMSINV (0.60) = 0.2534

Therefore, we get, P(-0.2534 < Z < 0.2534) = 0.20

For z = -0.2534

x1 = (-0.2534 * 0.9) + 9.7 = -0.2281 + 9.7 = 9.4719

x2 = (0.2534 * 0.9) + 9.7 = 0.2281 + 9.7 = 9.9281

Hence, we get,

Low = 9.4719

High = 9.9281