Statistics Chapter 6 Normal Distributions

Chapter 6 Normal Distributions

What is Normal Distribution?

Characteristics of Normal Distribution

 The curve is bell-shaped, with the highest point over the mean 𝜇  The curve is symmetrical about a vertical line through 𝜇  The curve approaches the horizontal axis but never touches or crosses it  The inflection (transition) points between cupping upward and downward occur above 𝜇 + 𝜎 𝑎𝑛𝑑 𝜇 − 𝜎

Let’s test your knowledge A) Do these distribution s have the same mean? B) Compare the two graphs, what can you tell me about the standard deviation?

Answer

 A) Both graphs have the same mean  B) Red curve has a smaller standard deviation because the data is more compressed. Black curve has a bigger standard deviation because the data is more spread out.

Empirical rule

 For a distribution that is symmetrical and bell-shaped (normal distribution):  Approximately 68% of the data values will lie within one standard deviation on each side of the mean

 Approximately 95% of the data values will lie within two standard deviations on each side of the mean  Approximately 99.7% of the data values will lie within three standard deviations on each side of the mean

Again look at the graph

Group Work

 Check Chebyshev’s theorem. What are the similarities and differences between them?

Potential Answer

 Empirical rule gives a stronger statement than Chebyshev’s theorem in it gives definite percentages, not just lower limits.

Group Work: Creating the normal distribution curve

 The yearly corn per acre on a particular farm is normally distributed with mean = 60 and standard deviation of 7. Create the normal distribution curve.

Answer

 49

56

63

60

67

74

81

Group Work: Empirical Rule

 The playing life of a Sunshine radio is normally distributed with mean 600 hours and standard deviation of 100 hours. What is the probability that a radio selected at random will last from 600 to 700 hours?

 34.1%

Technology

 If you have TI-83/TI-84  Press the Y= key. Then, under DISTR, select 1:normalpdf (𝑥, 𝜇, 𝜎) and fill in desired 𝜇 𝑎𝑛𝑑 𝜎 values. Press the WINDOW key. Set Xmin to 𝜇 − 3𝜎 𝑎𝑛𝑑 𝑋𝑚𝑎𝑥 𝑡𝑜 𝜇 + 3𝜎. Finally, press the ZOOM key and select option 0: ZoomFit

What are some applications of normal distribution curve?

 It gives us how reliable are the data

 Used for control charts

Control Charts

 It is useful when we are examining data over a period of equally spaced time interval or in some sequential order.

Example of Control Chart

How to Make a control chart of the random variable x  1) Find the mean and standard deviation of the x distribution by  A) Using past data from a period during which the process was “in control” or  B) using specified “target” values for 𝜇 𝑎𝑛𝑑 𝜎

 2) Create a graph in which the vertical axis represents x values and the horizontal axis represents time.  3) Draw a horizontal line at height 𝜇 𝑎𝑛𝑑 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙, 𝑑𝑎𝑠ℎ𝑒𝑑 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 − 𝑙𝑖𝑚𝑖𝑡 𝑙𝑖𝑛𝑒𝑠 𝑎𝑡 𝜇 ± 2𝜎 𝑎𝑛𝑑 𝜇 ± 3𝜎  4) Plot the variable x on the graph in time sequence order. Use line segments to connect the points in time sequence order

Example: Day 1 x

2

3

11 20 25

4

5

6

7

8

9

10 11 12 13 14 15

23

16

19

8

25

17

20

23

29

A) Find the mean and the standard deviation (population) B) Find control limit 𝜇 ± 2𝜎 C) Find control limit 𝜇 ± 3𝜎

18

14

10

Answer

 Mean = 18.53  Standard deviation = 5.82 𝜇 ± 2𝜎 = 6.89 𝑎𝑛𝑑 30.17 𝜇 ± 3𝜎 = 1.07 𝑎𝑛𝑑 35.99

Interpreting control chart

 Ms. Tamara of the Antlers Lodge examines the control chart for housekeeping. During the staff meeting, she makes recommendations about improving service or, if all is going well, she gives her staff a well-deserved “pat in the back”. Determine if the housekeeping process is out of control

Answer

 Graph to the left: there are 9 consecutive days on one side of the mean. It is a warning signal. It would mean that the mean is slowly drifting upward from 19.3.  Graph to the right: We have data value beyond 𝜇 + 3𝜎. We have two of three data values beyond 𝜇 − 2𝜎. Both are warning signals.

Homework Practice

 Pg 244 #1-7, 11,13

Standard Units and Areas Under the Standard Normal Distribution

Note:

 Normal distributions vary from one another in two ways: The mean 𝜇 may be located anywhere on the x axis, and the bell shape may be more or less spread according to the size of the standard deviation 𝜎

Scenario  Suppose you have two students comparing their test result in two different classes.  Jack got a 76 while the class average was a 66  Jill got an 82 while the class scored 72  Both scored 10 points above the mean.  How do we determine who did better in respect to their peers in the class?

The Answer

 You have to use the z test or z score because you want to know the position in term of the standard deviation away from the mean.

What is z test or z score?

 Z value or z score gives the number of standard deviations between the original measurement x and the mean 𝜇 of the x distribution

𝑧 =

𝑥−𝜇 𝜎

Note:

 Unless otherwise stated, in the remainder of the book we will take the word average to be either the sample arithmetic mean or population mean.

Calculating z scores  Suppose the company states that their average large pizza is 10 inches in diameter and the standard deviation is 0.7 inches. A customer ordered a pizza and found out it is of 7 inches. Assume the pizza follows a normal distribution. If the size of the pizza is below three standard deviation, the company would be in danger of getting customer complaints and have a bad name in its company.  How many standard deviation s from the mean is 7? Is the company going to be in trouble?

Answer

𝑧 =

𝑥−𝜇 𝜎

=

7−10 .7

= −4.28

It is 4.28 below the mean, and therefore the company is in trouble.

Group Work

 (Fake Data) National weigh average for men at 5’8 is 139 lbs with a standard deviation of 15.7. If you are 5’8 and weigh 171 lbs. Find the z score. Should you be concerned?

Answer

 𝑧=

𝑥−𝜇 𝜎

=

171−139 15.7

= 2.03

 Answer may vary depending how you think of it. If you consider outlier is when z=2.5 or more then no, but if you say that he is not with the 95% (within 2 standard deviation) then he should. (based on empirical rule)

Question:

 How would you find the raw score x if you are given the z score, mean and standard deviation?

Answer

𝑥 = 𝑧𝜎 + 𝜇

Group Work  Based on your experience, it takes you on average of 15 minutes to walk to school with standard deviation of 2 minutes.  A)On one particular day, it took you 12 minutes to get to school. What is the z score? Is the z value positive or negative? Why?  B) What would be the commuting time corresponding to a standard score of z=-2.5?

Answer

 A) z=-1.5. It is negative because it is shorter than the expected time.  B) 10 minutes

Standard normal distribution

 Standard normal distribution is a normal distribution with mean 𝜇 = 0 and standard deviation 𝜎 = 1

Area under the curve. The left tail rule

 Look at Appendix II Table 5 on A22

Group Activity  What is the value for and sketch the graph:  A) z=1.18  B) z=-2.35  C) z=0.56  D) z=-3.34

Group Activity: challenge  What’s the value for:  A) z>1.25  B) values in between z=1.21 and z=2.36  C) values to the right of 0.95  D) values in between z=-2.75 and z=1.38

Homework practice

 Pg 256 #1-48 eoe

Areas under any normal curve

How do you find the z score/z value?

Answer

 𝑧=

𝑥−𝜇 𝜎

Example:

 Let x have a normal distribution with 𝜇 = 7 𝑎𝑛𝑑 𝜎 = 3. Find the probability that an x value selected at random is between (5 and 8). 𝑃(5 ≤ 𝑥 ≤ 8)

Answer  You have to convert the x into z score:

 𝑍=

5−7 3

= −3

2

 𝑍=

8−7 3

=3

1

2

 𝑃 𝑧 = − 3 = .2546 1

 𝑃 𝑧 = 3 = .6293  .6293-.2546=.3747

Group Work

 A typical iphone have a battery life that is normally distributed with a mean of 18 hours with a standard deviation of 2 hours. What is the probability that an iphone will have a battery life that is greater than 24 hours?

Answer

 𝑃 𝑥 > 24 = 𝑃 𝑧 > 3 =.0013

Tech notes finding area under the curve in between z values

 For TI 83/TI 84

 Press DISTR key, select 2:normalcdf (𝑙𝑜𝑤𝑒𝑟 𝑏𝑜𝑢𝑛𝑑 𝑧 𝑣𝑎𝑙𝑢𝑒, 𝑢𝑝𝑝𝑒𝑟 𝑏𝑜𝑢𝑛𝑑 𝑧 𝑣𝑎𝑙𝑢𝑒, 𝜇, 𝜎)

Find x given probability

 A company sells a hybrid battery that has a mean life of 8 years with standard deviation of 2 years. How long can the company guarantee its warranty such that they do not want to refund on more than 5% of the hybrid battery?

Answer

 Since refund only occur when it falls below the mean. It is on the left side of the normal distribution curve. When is z such that it is .05?  Z=-1.645 𝑥−𝜇

 Using the formula 𝑧 = 𝜎 = −1.645 = 4.71, 4.71 𝑦𝑒𝑎𝑟𝑠. 𝑜𝑟 55 𝑚𝑜𝑛𝑡ℎ𝑠

𝑥−8 2

,𝑥 =

Group Work

 A company sells a product that has a mean shell life of 60 month with the standard deviation of 5 months. How long can the company guarantee its warranty such that they do not want to refund on more than 7% of the product?

Answer

 you have to use z=-1.48  -1.48*5+60=52.6 months

Example: Find z

 Find the z value such that 90% of the area under the standard normal curve lies between –z and z

Answer



1−.90 2

= .05, 𝑠𝑜 𝑧 = 1.645

 Z values between -1.645 and 1.645

Group work

 Find the z value such that 80% of the area under the standard normal curve lies between –z and z

Group work

 Find the z value such that 75% of the area under the standard normal curve lies between –z and z

Tech note for finding x value

 Example find an x value from a normal distribution with mean 40 and standard deviation 5 such that 97% of the area lies to left of x  TI 83/TI 84: press DISTR key and select 3:invNorm(𝑎𝑟𝑒𝑎, 𝜇, 𝜎)

Check for normality  How do we tell if data follow a normal distribution?

 1) Make a histogram, it should be roughly bell-shaped  2) Outliers. Make a box-and-whiskers plot (5 number summary)  Outliers are either greater than Q3+1.5(IQR) or below Q1-1.5(IQR)

 3) Skewness: Pearson’s index  𝑃𝑒𝑎𝑟𝑠𝑜𝑛′ 𝑠 𝑖𝑛𝑑𝑒𝑥 =

3 𝑥−𝑚𝑒𝑑𝑖𝑎𝑛 𝑠

,

 𝑛𝑜 𝑠𝑘𝑒𝑤𝑛𝑒𝑠𝑠 𝑖𝑓 𝑖𝑛𝑑𝑒𝑥 𝑖𝑠 𝑖𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 − 1 𝑎𝑛𝑑 1  Skewed distributions are not normal

4) Normal quantile plot (or normal probability plot) 

for TI 83/TI 84, put all the data in the calculator under edit, then turn on STATPLOT then highlight the sixth plot option

 It should look like a straight line

Check for normality 50

23

17

60

75

73

25

27

30

46

53

13

61

47

87

78

37

81

76

29

36

48

62

55

Homework Practice

 Pg 267 #1-30 eoe, 31

Normal Approximation to the Binomial Distribution

Normal approximation to the binomial distribution  Consider a binomial distribution where  𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠  𝑟 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠  𝑝 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠 𝑜𝑛 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑡𝑟𝑖𝑎𝑙  𝑞 = 1 − 𝑝 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 𝑜𝑛 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑡𝑟𝑖𝑎𝑙  If np>5 and nq>5, then r has a binomial distribution that is approximated by a normal distribution with

 𝜇 = 𝑛𝑝 𝑎𝑛𝑑 𝜎 = 𝑛𝑝𝑞  As n increases, the approximation becomes better

Example:

 Find the 𝑛, 𝑝, 𝑞, 𝜇, 𝜎 of the binomial distributions for which p=0.25, q=0.75, and the number of trials is first n=3, n=10, then n=25 and then n=50

Answer

 Find the 𝑛, 𝑝, 𝑞, 𝜇, 𝜎

Important note about binomial approximation

 Remember how to graph the binomial distribution? Remember how you have to incorporate the 0.5?

 For example when r=5, you draw the binomial distribution from 4.5 to 5.5.

How to make the Continuity Correction  Convert the discrete random variable r (number of successes) to the continuous normal random variable x by doing the following:  1) If r is a left point of an interval, subtract 0.5 to obtain the corresponding normal variable x; that is, x=r-0.5  2) If r is a right point of an interval, add 0.5 to obtain the correspoinding normal variable x: that is, x=r+0.5  For instance, 𝑃 6 ≤ 𝑟 ≤ 10 , where r is a binomial random variable is approximated by 𝑃(5.5 ≤ 𝑥 ≤ 10.5), where x is the corresponding normal random variable.

Normal Approximation  The owner of a new apartment building must install 25 water heaters. From past experience in other apartment buildings, she knows that Quick Hot is a good brand. A Quick hot heater is guaranteed for 5 years only, but from the owner’s past experience, she knows that the probability it will last 10 years is 0.25.

 A) What is the probability that 8 or more of the 25 water heaters will last at least 10 years?  B) How does this result compare with the result we can obtain by using the formula for the binomial probability distribution with n=25 and p=0.25?

 C) How do the results of part a and b compare?

Answer  A) since n=25 and p=0.25, r is the binomial random variable corresponding to the number of successes out of n=25. We want to find 𝑃(𝑟 ≥ 8).  𝜇 = 𝑛𝑝 = 25 ∗ 0.25 = 6.25  𝜎 = 𝑛𝑝𝑞 = 25 ∗ 0.25 ∗ 0.75 = 2.17  𝑠𝑖𝑛𝑐𝑒 𝑟 ≥ 8, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑥 ≥ 7.5

 𝑧=

𝑥−𝜇 𝜎

=

7.5−6.25 2.17

≈ 0.58 so, 1-0.7190=0.2810

 B) Using TI83/TI84 you get 0.2735  C) The error is only .0075, which is small.

Group Work:  The Denver Post stated that 80% of all new products introduced in grocery stores fail (are taken off the market) within 2 years. If a grocery store chain introduces 66 new products, what is the probability that within 2 years  A) 47 or more fail?  B) 58 of fewer fail?  C) 15 or more succeed?  D) fewer than 10 succeed?

Answer  𝑛𝑞 = 52.8>5, 𝑛𝑝 = 13.2>5, yes I can use normal approx to binomial distribution  1) 𝑛𝑞 = 52.8 𝜎 = 3.25, 𝑧 =

.0262=.9738  2) 𝑛𝑞 = 52.8 𝜎 = 3.25, 𝑧 =

46.5−52.8 =-1.94 3.25

= .0262, so 1-

58.5−52.8 =1.75= 3.25

 3) 𝑛𝑝 = 13.2, 𝜎 = 3.25, 𝑧 = .3446

14.5−13.2 3.25

 4) 𝑛𝑝 = 13.2, 𝜎 = 3.25, 𝑧 =

9.5−13.2 3.25

.9599

= 0.4 = .6554, 1 − .6554 =

= −1.14 = .1271

Group Work

 It is estimated that 3.5% of the general population will live past their 90th birthday. In a graduating class of 753 high school seniors, what is the probability that  A) 15 or more will live beyond their 90th birthday?  B) 30 or more will live beyond their 90th birthday?  C) Between 25 and 35 will live beyond their 90th birthday?  D) more than 40 will live beyond their 90th birthday?

Answer  𝑛𝑞 = 726.645>5, 𝑛𝑝 = 26.355>5, yes I can use normal approx to binomial distribution  1) 𝑛𝑝 = 26.355 𝜎 = 5.04, 𝑧 = .0094=.9906  2) 𝑛𝑝 = 26.355 𝜎 = 5.04, 𝑧 =

.7324=.2676  3)𝑛𝑝 = 26.355 𝜎 = 5.04, 𝑧 = 35.5−26.355 5.04

14.5−26.355 =-2.35=.0094 5.04 29.5−26.355 =.62=.7324 5.04

so, 1-

so, 1-

24.5−26.355 =-.37=.3557, 5.04

,𝑧 =

= 1.81 = .9649, 𝑠𝑜 .9649 − .3557 = .6092

 4)𝑛𝑝 = 26.355 𝜎 = 5.04, 𝑧 = .9975=.0025

40.5−26.355 =2.81=.9975=15.04

Use the examples and my answers to find out the pattern of the wording!

Homework Practice

 Pg 278 #1-14 eoo