Question

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 37 and a standard deviation of 9. Suppose that one individual is randomly chosen. Let X-percent of fat calories. Round all answers to 4 decimal places if where possible a. What is the distribution of X? X-N b. Find the probability that a randomly selected fat calorie percent is more than 41. c. Find the minimum number for the upper quarter of percent of fat calories. Question Help: Read Video Video Message instructor

Answer 1

Solution :

Given that ,

a) The distribution of X is normal X $\sim$ N ( 37, 9 )

b) P(x > 41) = 1 - p( x< 41)

=1- p P[(x - $\mu$) / $\sigma$ < (41 - 37) / 9]

=1- P(z < 0.44)

= 1 - 0.6700

= 0.3300

c) The z dist'n Third quartile is,

= P(Z < 0.674 ) = 0.75

z = 0.674

Using z-score formula,

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

x = 0.674 * 9 + 37

x = 43.066

Third quartile =Q3 = 43.07