The Empirical Rule
Z Scores (Standardized Score)
How many standard deviations away from the mean your value x is
Using the Normal Table
- A study of college freshmen's study habits found that the time (in hours) that college freshmen use to study each week follows a normal distribution with a mean of 7.2 hours and a standard deviation of 5.3 hours
How many hours do the students who study in the top 15% spend studying?
The middle 68%?
Top 15%: 12.5 hours
Middle 68%: 1.9 hours to 12.5 hours
- Suppose that the weight of navel oranges is normally distributed with mean of 8 ounces, and standard deviation of 1.3 ounces. And the weights of Valencia oranges is normally distributed with mean of 9 ounces, and standard deviation of 1.6 ounces
You grow a navel orange that weighs 9.5 ounces and a Valencia orange that weight 10.5 ounces, which should you enter in the giant fruit contest?
Z score for navel orange = (9.5-8)/1.3 = 1.1538
Z score for Valencia orange = (10.5-9)/1.6 = 0.9375
The weights of newborn children in the United States vary according to the Normal Distribution with mean 7.5 pounds and standard deviation 1.25 pounds.
What is the probability that a baby chosen at random weighs less than 5.5 pounds at birth?
- Draw a sketch
Calculate Z score
Look up probability on the normal table
The composite score of students on the ACT college entrance examination in a recent year had a Normal distribution with mean of 20.4 and standard deviation of 5.8
What is the probability that a randomly chosen students scored 24 or higher on the ACT?
2ND + VARS (DISTR) ➡️ 2: normalcdf
Normalcdf(lower, upper, mean, standard deviation)
- What is the probability that a randomly chosen student scored between a 19 and a 24 on the ACT?
What score would someone in the 15th percentile have scored?
Find the z value on the normal table
Solve for x
Calculator: invNorm(area, mean, standard deviation)
![ф TEns [NSTRUMENTS 14 „ гоямдтп си: ТАТЕ ](./media/image157.png)
Suppose that the mean height of men is 70 inches with a standard deviation of 3 inches. And suppose that the mean height for women is 65 inches with a standard deviation of 2.5 inches
If the heights of men and women are Normally distributed, find the probability that a randomly selected woman is taller than a randomly selected man.
- Find the necessary information
|W-M||65-70 = -5||Sqrt(3^2+2.5^2) = 3.9|
Calculate Z score
Find the probability on the table and subtract that from 1
1-0.9015 = 0.0985 = 9.85%
Suppose that the height (X) in inches, of adult men is a normal random variable with mean of 70 inches. If P (X>79) = 0.025
What is the standard deviation of this random normal variable?
Find the z score on the calculator: invNorm(area)
![ТИ-84 PIus S'tver Editj(n ф TEns [NSTRUMENTS гоямдта ](./media/image161.png)
Solve for SD