In: Economics
The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 40 and a standard deviation of 15 . Suppose that one individual is randomly chosen. Let X= percent of fat calories.
In each appropriate box you are to enter either a rational number in "p/q" format or a decimal value accurate to the nearest 0.01 .
a).
Consider the given problem here “X” be a random variable denotes “percent of fat calories”. Now, “X” follows normal distribution with mean “40” and standard deviation “15”. So, if we pick any individual then the probability distribution must be same, => “xi” also follows the normal distribution with mean “40” and standard deviation “15”.
c).
The probability that a person consume more than 40 percent fat is given by.
=> P[X > 40] = 1 - P[X < 40] = 0.5, as we know that normal distribution is symmetrical about its mean and here the mean value is “40”, => probability “more than 40” and “less than 40” are same that is “0.5”. So, the required probability is “0.5”.
e).
Let’s assume “t” be standard normal variable having mean “0” and standard deviation “1”. So, here the value of “Q3” is given by, “Q3=0.67” (from the standard normal table). Since, for “t < 0.67” the probability is approximately “0.75”.
So, the lower quartile of "t" is given by, “Q1 = (-0.67)”, since the standard normal distribution is symmetrical about its mean zero. So, the lower quartile for “X” is given by “Q1*15 + 40 = (-0.67)*15+40 = 29.95, where "40" be the mean and "15" be the standard deviation of "X".
g).
Now, in “part-c” the required probability is given by, “P = (X > 40)”, => if we pick any person then there is 50% chance that the particular person consume calories more than “40%”.