Download Z-scores and Empirical Rule Notes

Z-scores and Empirical Rule Notes
Name: _______________________
Empirical Rule says that …
68% of the data in a normally distributed data set is within 1 standard deviation.
95% of the data in a normally distributed data set is within 2 standard deviations.
99.7% of the data in a normally distributed data set is within 3 standard deviations.
Types of questions involving the Empirical Rule:
1. The scores on a university examination are normally distributed with a mean of 70 and a
standard deviation of 10. If the middle 68% of students will get a “C”, what is the lowest
mark that a student can have and still be awarded a C?
To solve: The middle 68% of students are within 1 standard deviation of the mean
according to the Empirical Rule. The question wants to know the LOWEST mark
that a student can get o receive a C, so you must subtract 1 standard deviation from
the mean.
70 – 10= 60
2. The lifetime of lightbulbs of a particular type are normally distributed with a mean of 100
mmHg and a standard deviation of 6 mmHg. What percentage of 18-year-old women have
a systolic blood pressure between 88 mmHg and 112 mmHg?
To solve: You must decide how many standard deviations away each of the given blood
pressures are from the mean. Start by looking at the left of the mean. 100 – 6 = 94.
That’s not far enough, so you subtract another standard deviation. 94 – 6 = 88.
Because 88 is two standard deviations away from the mean, 95% of 18-year-old
women have a systolic blood pressure between 88 mmHg and 112 mmHg.
Z-scores:
xx

Z-scores are used to normalize data, or to convert all data to a common unit. A z-score tells
you how many standard deviations away your data is from the mean.
Types of z-score questions:
1. Lewis earned 80 on his biology midterm and a 71 on his history midterm. In the biology
class the mean score was 75 with a standard deviation of 4. In the history class the mean
score was 73 with a standard deviation of 3.
a. Convert each score to a standard z-score.
To solve: Biology:
80  75
 1.25
4
History:
71  73
 0.67
3
b. On which test did he do better compared to the rest of the class?
Solution: In both classes he did worse than the average because both z-scores were negative.
However, he did better compared to the rest of the class in History because his z
score is smaller.
Z-table:
The z-table gives you the area, probably, or percent of data that is below the said value.
You use the z-table when you see one of the above bold words.
How to use the z-table: You need to have one number before the decimal, and two numbers
after the decimal.
 If there is no number before the decimal, put a 0 before the decimal.
 If there is only one number after the decimal, add a 0 on the end.
 If there is no decimal, add one then add two 0s after it.
 If there are more than two numbers after the decimal you have to ROUND.
o Look at the 4th number after the decimal. If it is a 4 or below, keep the 3rd
number the same.
 Example: 1.45345698 becomes 1.45 because 3 is the 4th number.
o Look at the 4th number after the decimal. If it is a 5 or above, raise the 3rd
number 1 higher than it was before.
 Example: -0.86795643 becomes -0.87 because 7 is the 4th number.
Types of z-table questions:
A class of 217 students participated in a softball throw for the distance test. The mean
performance of the group was 173 and the standard deviation was 31. Based on this data,
answer the following questions:
a. What percentage of students was able to throw the softball between 151 and 180?
To solve: Because the questions asks for the percentage, you must use your z-table. In
order to use your z-table, you must convert your throw values to z-scores.
151  173
180  173
 0.71 when rounded
 0.23 when rounded
31
31
Next find both of these z-scores on the z-table: 0.2389 and 0.5910
To find the percentage between two numbers you subtract the lower from the higher:
0.5910 – 0.2389 = 0.3521 = 35.21%
b. What percentage of the students could throw farther than 200 feet?
200  173
 0.87 . Then find the z-table value: 0.8078
To solve: Find the z-score first:
31
To find the percentage of students who throw farther than 200, you must subtract
your z-table value from 1. 1 – 0.8078 = 0.1922 = 19.22%
c. What percentage of the students could only throw less than 114 feet?
To solve: The z-table values give you the percent that throws less than the given amount.
Therefore, once you find the z-score, simply look it up on your z-table.
114  173
 1.90 Then find the z-table value: 0.0287 = 2.87%
Z-score:
31
Normal Distribution:
A normal distribution looks like a bell curve. In order to use the Empirical Rule or a Z-table
your data must be normally distributed.
Use what you know about a normal distribution to answer the following questions:
1. Which graph above has a larger mean?
Solution: Graph B has a larger mean because the mean is located in the middle of each
graph and the mean of graph B is located further to the right.
2. Which graph has a larger standard deviation?
Solution: Graph B has a larger standard deviation because it is more spread apart.
Exercises:
1. What percent of data is within 1 std deviation of the mean?
2. What percent of data is within 2 std deviations of mean?
3. What percent of data is within 3 std deviations of mean?
4. The scores on a university examination are normally distributed with a mean of 62 and a
standard deviation of 11. If the middle 68% of students will get a “C”, what is the lowest
mark that a student can have and still be awarded a C?
5. The lifetimes of lightbulbs of a particular type are normally distributed with a mean of
360 hours and a standard deviation of 5 hours. What percentage of the bulbs have lifetimes
that lie within 2 standard deviation of the mean?
A) 31%
B) 84%
C) 68%
D) 95%
6. The systolic blood pressure of 18-year-old women is normally distributed with a mean of
120 mmHg and a standard deviation of 12 mmHg. What percentage of 18-year-old women
have a systolic blood pressure between 96 mmHg and 144 mmHg?
A) 99%
B) 68%
C) 95%
D) 99.99%
7. Lewis earned 85 on his biology midterm and 81 on his history midterm. In the biology
class the mean score was 79 with a standard deviation of 5. In the history class the mean
score was 76 with a standard deviation of 3.
(a) Convert each score to a standard z score.
(b) On which test did he do better compared to the rest of the class?
4. On one measure of attractiveness, scores are normally distributed with a mean of 3.93
and a standard deviation of .75. Find the probability of randomly selecting a subject with
a measure of attractiveness that is greater than 2.75.
5. The serum cholesterol levels in men aged 18 to 24 are normally distributed with a mean
of 178.1 and a standard deviation of 40.7. If a man aged 18 to 24 is randomly selected, find
the probability that his serum cholesterol level is between 100 and 200.