Chapter 6:

I WANT YOU, TO ACE THE FINAL

Normal Probability Distribution

Dan, Jenna, Jess, Mauri, Reid, and Tanner

Introduction In past chapters we covered different types of data distribution. In chapter 6, we moved from discrete probability distributions to continuous probability distributions. Although part of the chapter focuses on uniform distributions, where the probabilities for all events are the same, most of the chapter focuses on normal distribution. Normal distribution is probability expressed by a bell shaped curve with the most probable value being the mean value.

6.2 Vocab and Formulas Normal Distribution- If a continuous random variable has a distribution with a graph that is symmetrical and bell-shaped it can be described as normal distribution. Uniform Distribution- A continuous random variable has uniform distribution if if its values spread evenly over the range of possibilities. It results in a rectangular shape. Density Curve- A graph of continuous probability Requirements: - must have an area equal to 1 - every point on the curve must have a vertical height that is 0 or greater and thus cannot fall below the x axis Standard Normal Distribution- A normal probability distribution with a mean of 0 and a standard deviation of 1. Also total area under density curve is equal to one

6.3 Formulas Key concept: Working with normal distributions that are not standard.

Formula for Z score: Used to convert from the nonstandard normal distribution to the standard normal distribution.

Example Assume the weight of men are normally distributed with a mean of 172 lbs and standard deviation of 29 pounds. Find the probability of selecting one man and having him weigh less than 174 lbs. 1) Draw a bell shaped curve shaded left of 174 lbs. 2) Use formula 6-2: z= 174-172 = 0.07 29 3) Using the z=0.07, refer to the table to find the cumulative area is 0.5279

6.4 Vocab and Formulas Sampling distribution of a statistic- the distribution of all values of the statistic when all possible samples of the same size “n” are taken from the same population. Sampling distribution of the proportion- the distribution of sample proportions, with all samples having the same sample size “n” taken from the same population. Sampling distribution of the mean- is the distribution of sample means, with all samples having the same size “n” taken from the same population. Sampling variability- the value of a statistic, such as the sample mean x, depends on the particular values include in the sample, and it generally varies from sample to sample. P(x)= 1 2(2-2x)!(2x)!

Formula to describe a Sampling Distribution

6.4 Example A quarterback threw 1 interception in his first game, 2 interceptions in his second game, 5 interceptions in his third game, and then retired. Consider the population consisting of values 1,2,5. Note that two of the values (1 and 5) are odd, so the proportion of odd numbers in the population is ⅔.

Sample

Proportion of odd numbers

Probability

1,1

1

1/9

1,2

.5

1/9

1,5

1

1/9

2,1

.5

1/9

2,2

0

1/9

2,5

.5

1/9

5,1

1

1/9

5,2

.5

1/9

5,5

1

1/9

Section 6.5: Central Limit Theorem Difference between 6.4 and 6.5 In Section 6.4, the n was always 1. In Section 6.5, we learned how to find the value of x, z-score, or percentage with an n greater than 1. Essential Principles (May be helpful later… but we don’t have to worry about it now) 1. n greater than 30 than distribution can be approximated. 2. n less than or equal to 30 and has a normal distribution, then mean and standard deviation is normal. 3. n less than or equal to 30 the original population does not have a normal distribution, methods do not apply.

z= x - µ Ơ n

* When working with an individual value from normally distributed population, use the methods of Section 6.3 and 6.4. Use the original formula. * When working with a mean for some sample (greater than 1), be sure to divide the the original z-score formula by the square root of n.

Example 6.5 Assume that women’s heights are normally distributed. Mean= 63.6 Standard Deviation= 2.5 in. If 100 women are randomly selected, find the probability that they have a mean height greater than 63 in.

z= (63) - (63.6) (2.5) 100

z= -2.4 10

z= x - µ Ơ n z= -0.024

Then, use chart in the back of the book 0.4920

Quiz 1) Section 6.2 At a local gym, the length of yoga classes are uniformly distributed between 40 and 42 minutes. Find the probability that a randomly selected class will last longer than 41 minutes.

#1 Solution P(x > 41) = Area Area = length x width Area = .5 x (42-41) Area = .5 = Probability

#2 2) Section 6.3 A group of high school students has a mean age of 15.8 years with a standard deviation of .6 years. What percentage of students are younger than 16 years old?

#2 Solution μ = 15. 8 σ = .6 x = 16 z = x - μ = 16 - 15.8 = .33 σ .6 Look at the z-score chart in the back of the book for .33 and see that it correlates to .6293 .6293 = 62.93%

#3 3) Section 6.3 Based on the results from the last example, what percent of high school students are between 15.8 and 16 years old?

#3 Solution 3) 16 year olds = .6293 15.8 year olds = .5000 (Because 15.8 is also the mean, the probability will always be .5000) Between = .6293 - .5000 .1293 12.93%

#4 4) Section 6.4 A group of students at LCC is randomly selected and asked how many animals they each have. One student had one animal, another had 2 animals, and the third had 5 animals. Consider the populations to consist of the values 1,2, and 5. List all the different possible samples of size n=2 with replacement. Find the mean, probability, and mean of the sampling distribution. Does the mean of the sampling distribution equal the mean of the population?

#4 Solution

#5 5) Section 6.5 The average GPA at a particular school is 2.89 (mean = 2.89) with a standard deviation of s=0.63. A random sample of 25 students is collected. Find the probability that the average GPA for this sample is greater than 3.0.

#5 Solution