Homework SourseNormal Probability Distributions 6895997 Rule I The serum to
Normal Probability Distributions
68-95-99.7 Rule
I. The serum total cholesterol for males 20 to 29 years old is approximately normally distributed with mean 180 and standard deviation 36.2, based on data obtained from the National Health and Nutrition Examination Survey. Use the 68-95-99.7 rule to determine:
The middle 68% of values. What percent of values fall below this range? What percent fall above this range?
What range of values will capture the middle 95%
Standardizing a normal random variable: Find the Z-score of any particular value
What is a z score for a serum total cholesterol level of 170?
What is a z score for a serum total cholesterol level of 190?
A continuous random variable, x, is normally distributed with a mean of 200 grams and a standard deviation of 25 grams. Convert each of the following x values into its corresponding z-scores:
x= 150 d. x=180
x= 200 e. x = 285
x= 315
II. A pediatrician obtains the heights of her 200 three-year-old female patients. The heights are approximately normally distributed, with mean 38.7 inches and standard deviation3.17 inches. Use the normal model to determine the proportion of the 3-year-old females that have a height less than 35 inches.
Find the probability that a randomly selected 3-year-old female is between 35 and 40 inches tall, inclusive.
I. It has been reported that the average monthly cell phone is $50. Assuming a normal distribution and a standard deviation of $10, what is the probability that randomly selected cell phone subscriber’s bill last month was less than $35? More than $70?
II. It has been reported that households in the West spend an annual average of $6050 for groceries. Assume a normal distribution with a standard deviation of $1500.
What is the probability that a randomly selected household spends more than $6350 for groceries?
How much money would a Western household have to spend on groceries per year in order to be at the 99th percentile (i.e., only 1 percent of Western households would spend more on groceries)?
Solution
a) The middle 68% of values.
= (180 + / - 36.20)
= 143.8 to 216.2 are middle 68 percent of the values.
What percent of values fall below this range?
= (100 - 68 ) / 2
= 34 /2
= 17%
What percent fall above this range?
= (100 - 68 ) / 2
= 34 /2
= 17%
Middle 95% of the values are:
= 180 +/- (2*36.2)
= 180 +/- 72.4
= (107.6, 252.4)
b) Z-score = X - mean / S.D
c)
Mean = 38.7
S.d = 3.17
P( X < 35) = P ( Z < 35 - 38.7 / 3.17/sqrt(200) )
= P ( z < -16.50)
= 0
P ( 35 < X < 40)
= 0.999
d)
Mean = 50
Sd = 10
P (X < 35) = P (Z < 35 - 50 /10)
= P ( Z < -1.5)
= 0.0668
P ( Z > 70)
= 1 - P (Z < 70)
= 1 - 0.9772
= 0.02275
Hope these help
| Mean= | 200 | SD= | 25 |
| X | z-score | ||
| 150 | -2 | ||
| 200 | 0 | ||
| 315 | 4.6 | ||
| 180 | -0.8 | ||
| 285 | 3.4 |
