6.2: The normal distribution

Transcription

1 6.2: The normal distribution

2 Normal Distributions Characterized by symmetric, bell-shaped (mound-shaped) curve. Heights, weights, standardized test scores A particular normal distribution is determined by The mean µ The standard deviation

3 Normal Distribution and Deviation from the Mean

4 Example: Adult Heights 95% of female adult heights are between 58 and 72 inches 95% of male adult heights are between 62 and 78 inches

5 Z-scores (revisited) The multiples 1, 2, and 3 or the number of standard deviations from the mean are denoted by z. For a particular observation, x, its z-score is computed by z = x µ For each fixed number z, the probability within z standard deviations of the mean is the area under the normal curve between and µ z µ + z

6 Finding Probabilities for the Normal Distribution What if we want the probability within 1.43 standard deviations of the mean? For normal distributions there is a table we can use (Table A in back of the book). It tabulates the normal cumulative probability falling below the point µ + z

7 Finding Probabilities for the Normal Distribution: P(-1.43< z < 1.43) To find P(-1.43 < z < 1.43) we do it in 3 steps: 1) We first find P(z < 1.43), using Table A or calculator 2) We then also know P(z > 1.43), which by symmetry means we know P(z < -1.43). 3) P(1.43 < z < 1.43) = P(z < 1.43) - P(z < -1.43)

8 To use Table A: Find the corresponding z-score. Look up the closest standardized score (z) in the table. First column gives z to the first decimal place. First row gives the second decimal place of z. The corresponding probability found in the body of the table gives the probability of falling below the z-score.

9 Part of Table A

10 The Probability Less Than 1.43 Standard Deviations

11 P(height < 70)=P(z<1.43) =

12 Example 1: Mensa Mensa is a society of high-iq people with IQ test scores at the 98 th percentile or higher. The Stanford- Binet IQ test scores that are used for admission are approximately normally distributed with a mean of 100 and a standard deviation of 16. How many standard deviations above the mean is the 98 th percentile? What is the IQ score for that percentile?

13 Example 1: Mensa Solution

14 Example 1: Mensa Solution 98 th percentile corresponds to a z- score of 2.05 So a person needs an IQ score of at least (16) = 133.

15 Example 2: SAT and ACT scores SAT and ACT exams are the two primary college entrance exams. Both have a mathematics component. The scores for the SAT range from 200 to 800 and are normally distributed with a mean of 500 and a standard deviation of 100. The scores for the ACT range from 1 to 36 and are normally distributed with a mean of 21 and a standard deviation of 4.7. u Which is better, a 650 on the SAT or a 30 on the ACT? We will answer by looking at percentiles. Begin by finding z-scores!

16 Example 2: SAT and ACT Solution 650 on the SAT (mean is 500, std. dev. is 100)

17 Example 2: SAT and ACT Solution 650 on the SAT (mean is 500, std. dev. is 100) z-score is ( )/100 = From Table A, this is in the 93 rd percentile. In other words, 7% of people scored above 650.

18 Example 2: SAT and ACT Solution 650 on the SAT (mean is 500, std. dev. is 100) z-score is ( )/100 = From Table A, this is in the 93 rd percentile. In other words, 7% of people scored above on the ACT (mean is 21, std. dev. is 4.7) z-score is (30-21)/4.7 = From Table A, this is in the 97 th percentile. In other words, 3% of people scored above 30.

19 Example 2: SAT and ACT Solution 650 on the SAT (mean is 500, std. dev. is 100) z-score is ( )/100 = From Table A, this is in the 93 rd percentile. In other words, 7% of people scored above on the ACT (mean is 21, std. dev. is 4.7) z-score is (30-21)/4.7 = From Table A, this is in the 97 th percentile. In other words, 3% of people scored above 30. Thus, a 30 on the ACT is better than a 650 on the SAT

20 Finding Probabilities on TI-83/84 Normalcdf(low,high,mean,std. dev.) For calculating P(a < X < b) when X has a normal distribution of mean mu and standard deviation sigma. On Calculator: 2 nd DISTR 2 a,b,mu,sigma) ENTER Normalcdf Invnorm(% to left,mean,std. dev.) For finding value a so that P(X a) = p, when X has normal distribution of mean mu and standard deviation sigma. On Calculator: 2 nd DISTR 3 p,mu,sigma) ENTER Invnorm

21 Finding Probabilities on TI-83/84 Normalcdf(low,high,mean,std. dev.) What percent of women are between 65 and 70 inches? On Calculator: 2 nd DISTR 2 65,70,65,3.5) ENTER Invnorm(% to left,mean,std. dev.) How tall does a woman need to be to be in the top 10%? On Calculator: 2 nd DISTR 3.9,65,3.5) ENTER

22 Finding Probabilities on TI-83/84 Normalcdf(low,high,mean,std. dev.) What percent of women are between 65 and 70 inches? Answer: 42.34% Invnorm(% to left,mean,std. dev.) How tall does a woman need to be to be in the top 10%? Answer: 69.5 inches, or 5 9.5

23 Building an Interval that Contains a Certain Percentage of the Data Suppose we have a normal distribution. We want the interval that contains 95% of the data (in terms of z values, i.e., between z* and z*). The Emperical Rule told us about 2 standard deviations but we want to be more precise. This means that 5% of the data must not be between, and of this amount 2.5% will be to the left of z*. Since 2.5% is , we look in Table A for a z-score with an entry of This gives us We conclude that 95% of the data lies between and 1.96.

24 Building an Interval that Contains a Certain Percentage of the Data (cont.) For normal distributions, 95% of the data has z-score between and Recall that female adult heights are normally distributed with a mean of 65 inches and a standard deviation of 3.5 inches. We can convert the z-scores into heights. We conclude that 95% of adult women have a height between inches and inches.

25 Unusual Observations Adult male heights are normally distributed with a mean of 70 inches and a standard deviation of 4 inches. Consider these two Sam is 79 inches tall (z-score is 2.25; corresponds to in Table A) Joe is 61 inches tall (z-score is -2.25; corresponds to in Table A) For a given person, we can think of unusual in two ways Sam is unusually tall, he is in the rarest 1.22% of tall people. Joe is unusually short, he is in the rarest 1.22% of short people. Both have unusual height, they are in the rarest 2.44%

26 P-Values The P-value is a measure of just how unusual the data is, in terms of what percentage of the data is even more unusual than the given data. Recall that Sam is unusually tall, he is in the rarest 1.22% of tall people. Joe is unusually short, he is in the rarest 1.22% of short people. Both have unusual height, they are in the rarest 2.44% This can be restated as Sam s one-tail (right-tail) P-value is Joe s one-tail (left-tail) P-value is Both have a two-tail P-value of

27 Graphical Depiction of P-Values

28 Other Types of Distributions We will also work with other distributions. Some will not be symmetric. For a distribution like this, we are only interested in one-tail (right-tail) P-values. Commuting time of 45 minutes has a P-value of 0.15.

29 Finding Probabilities on TI-83/84 Normalcdf(low,high) What is the percentage of data between 1 and 1.75 standard deviations? Normalcdf(1,1.75) Normalcdf(low,high,mean,std. dev.) What is the percentage of women between 62 and 70 inches? Normalcdf(62,70,65,3.5) Invnorm(% to left) What is the z-score for data in the top 10%? Invnorm(0.9) Invnorm(% to left,mean,std. dev.) How tall does a woman need to be to be in the top 10%? Invnorm(0.9,65,3.5)