# The Empirical Rule

# Z Scores (Standardized Score)

How many standard deviations away from the mean your value x is

# Using the Normal Table

# Practice Questions

- 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?

- Sketch

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?

Sketch

Find the z value on the normal table

Z≈-1.035

Solve for x

Calculator: invNorm(area, mean, standard deviation)

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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.

- Sketch

- Find the necessary information

Mean | SD | |
---|---|---|

Men | 70 | 3 |

Women | 65 | 2.5 |

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?

Sketch

Find the z score on the calculator: invNorm(area)

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Solve for SD