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 8.8 and 9.9 grams of sodium per day. C. The middle 30% of American men consume between what two weights of sodium?

Answer 1

Answer:

(a) The distribution of X is N (9.8, 0.8²).

(b) The probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c) The middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

Step-by-step explanation:

The random variable X is defined as the amount of sodium consumed.

The random variable X has an average value of, μ = 9.8 grams.

The standard deviation of X is, σ = 0.8 grams.

(a)

It is provided that the sodium consumption of American men is normally distributed.

The random variable X follows a normal distribution with parameters μ = 9.8 grams and σ = 0.8 grams.

Thus, the distribution of X is N (9.8, 0.8²).

(b)

If X ~ N (µ, σ²), then $Z=\frac{X-\mu}{\sigma}$, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z ~ N (0, 1).

To compute the probability of  Normal distribution it is better to first convert the raw score (X) to z-scores.

Compute the probability that an American consumes between 8.8 and 9.9 grams of sodium per day as follows:

$P(8.8$

                           $=P(-1.25$

Thus, the probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c)

The probability representing the middle 30% of American men consuming sodium between two weights is:

$P(x_{1}$

Compute the value of z as follows:

$P(-z$

The value of z for P (Z < z) = 0.65 is 0.39.

Compute the value of x₁ and x₂ as follows:

$-z=\frac{x_{1}-\mu}{\sigma}\\-0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8-(0.39\times 0.8)\\x_{1}=9.488\\x_{1}\approx9.5$     $z=\frac{x_{2}-\mu}{\sigma}\\0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8+(0.39\times 0.8)\\x_{1}=10.112\\x_{1}\approx10.1$

Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.