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 1 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 12.1 grams of sodium per day.

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

Answer 1

A)

The distribution of X is N(9.8, 1)

B)

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 9.8

S.d = 1

P(10.4<x<12.1) = P(x<12.1) - P(x<10.4)

P(x<12.1)

Z = (12.1-9.8)/1 = 2.3

From z table, P(z<2.3) = 0.9893

P(x<10.4)

Z = (10.4 - 9.8)/1 = 0.6

From z table, P(z<0.6) = 0.7257

Required probability is 0.9893 - 0.7257 = 0.2636

C)

We need to find 20% confidence interval

Critical value z from z table, For 20% interval is 0.25

Margin of error (MOE) = Z*S.D = 0.25*1 = 0.25

low = mean - moe = 9.8-0.25 = 9.55

High = mean + moe = 9.8+0.25 = 10.05