Question

The average American man consumes 9.6 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 9.5 and 10.1 grams of sodium per day.

c. The middle 10% of American men consume between what two weights of sodium?
Low:
High:

Answer 1

Part a)

X ~ N ( µ = 9.6 , σ² = 0.64 )

Part b)

X ~ N ( µ = 9.6 , σ = 0.8 )
P ( 9.5 < X < 10.1 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 9.5 - 9.6 ) / 0.8
Z = -0.125
Z = ( 10.1 - 9.6 ) / 0.8
Z = 0.625
P ( -0.12 < Z < 0.62 )
P ( 9.5 < X < 10.1 ) = P ( Z < 0.62 ) - P ( Z < -0.12 )
P ( 9.5 < X < 10.1 ) = 0.734 - 0.4503
P ( 9.5 < X < 10.1 ) = 0.2838

Part c)
X ~ N ( µ = 9.6 , σ = 0.8 )
P ( a < X < b ) = 0.1
Dividing the area 0.1 in two parts we get 0.1/2 = 0.05
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.05
Area above the mean is b = 0.5 + 0.05
Looking for the probability 0.45 in standard normal table to calculate critical value Z = -0.1257
Looking for the probability 0.55 in standard normal table to calculate critical value Z = 0.1257
Z = ( X - µ ) / σ
-0.1257 = ( X - 9.6 ) / 0.8
a = 9.4994
0.1257 = ( X - 9.6 ) / 0.8
b = 9.7006
P ( 9.4994 < X < 9.7006 ) = 0.1