Chapter 7 7.1 The Standard Normal Curve Introduction Probability density curve: The area under a probability density curve between any two values a and b has two interpretations: 1. 2. The region above a single point has zero width. For this reason, if X is a continuous random variable, then P(X = a) is equal to, and P(a < X < b) = P(a X b) for any number a and b. For any probability density curve, the area under the entire curve is equal to, because this area represents the entire population. : Interpret the area under a probability density curve Voltages in a circuit vary between 6 and 12 volts and are spread evenly over a range of possibilities (i.e., they have a uniform distribution - see picture). a. Find the probability of getting a voltage less than 11 volts. b. Find the probability of getting between 6.5 and 8 volts. c. Find the probability of getting anything not between 6.5 and 8 volts. Recall: The Rule of Complements says that P(not A) =.
The Normal Distribution Probability density curves come in many varieties, depending on the characteristics of the populations they represent. Many important statistical procedures can be carried out using only One type of probability density curve, called a. A population that is represented by a normal curve is said to be, or to have a. Figure 7.4 presents some examples of normal curves. The location and shape of a normal curve reflect the and of the population. The curve is symmetric around its. Therefore, the mode is equal to the. The population measures the spread of the population. Therefore, the normal curve is when the population standard deviation is, and when the population standard deviation is. Properties of Normal Distributions 1. Normal distributions have 2. Normal distributions are 3. The mean and median of a normal distribution are both equal to
4. The normal distribution follows the Empirical Rule: Finding Areas Under the Standard Normal Curve A normal distribution can have any and any positive, but it is only necessary to work with the normal curve that has mean and standard deviation. The probability density function for the standard normal distribution is called the. When finding an area under the standard normal curve, we use the letter z to indicate a value on the horizontal axis (see Figure 7.6). We refer to such a value as a. Since the mean of the standard normal distribution, which is located at the mode, is 0, the z-score at the mode of the curve is. Points on the horizontal axis to the left of the mode have, and points to the right of the mode have.
: Finding an area to the left of a z-score Use Table A.2 to find the area to the left of z = -1.63. Solution Step 1: Step 2: : Finding an area to the right of a z-score Use Table A.2 to find the area to the right of z = 1.32 Solution Step 1: Step 2: : Finding an area between two z-scores Use Table A.2 to find the area between z = -0.32 and z = 1.15 Solution Step 1: Step 2:
Step 3: Now, then: how about a much better way. Calculator! a)find the area to the right of z = 1.15 b)find the area to the left of z = 2.31 c)find the area between z = -2 and z = 2. What should this area basically be? Why? Finding a z-score Corresponding to a Given Area Often we are given an area and we need to find the z-score that corresponds to an area under the standard normal curve. This can be done using either Table A.2 or technology. Recall, the mode, z = 0 has an area of to its right and left. a)find the z score that has an area of 0.26 to its left. b)find the z score that has an area of 0.02 to its right.
c)find the z score that has an area of 0.26 to its right. d)find the z score that has an area of 0.02 to its left. Definition: The notation z α Let α be any number between 0 and 1; ion other words, 0 < α < 1. The notation z α refers to the z-score with an area of α to its right. a)find z 0.32 b)find z 0.84 Do You Know How to use a probability density curve to describe a population? How to use a normal curve to describe a normal population? How to find areas under the standard normal curve? How to find z-scores corresponding to areas under the normal curve? 7.2 Applications of the Normal Distribution Introduction In Section 7.1, we found areas under a standard normal curve, which has mean and standard deviation. We will learn to find areas under normal curves with any mean and standard deviation.
Converting Normal Values to z-scores Let x be a value from a normal distribution with mean µ and standard deviation σ. We can convert x to a z-score by using a method known as. Definition Let x be a value from a normal distribution with mean µ and standard deviation σ. The z-score of x is z = Properties of the z-score 1. 2. 3. Because the z-score follows a standard normal distribution, we can use the methods of Section 7.1 to find areas under any normal curve, by standardizing the original value. Rounding Off z-scores The z-score in Table A.2 are expressed to two decimal places. For this reason, when converting normal values to z-scores, we will round off the z-scores to two decimal places. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. a)what proportion of people have an IQ that is greater than 131.5? b)what proportion of people have an IQ that is between 110 and 120. c)find the 80 th percentile IQ score.
Recall: An area under a normal curve over an interval can be interpreted in two ways: 1. 2. The weight of male babies less than 2 months old in the United States is normally distributed with a mean of 11.5 pounds and a standard deviation of 2.7 pounds. a)what proportion of babies weigh more than 13 pounds? b)what proportion of babies weigh less than 15 pounds? c)what proportion of babies weigh between 10 and 14 pounds? d)is it unusual for a baby to weigh more than 17 pounds? e)find the 81 st percentile of the baby weights. f)find the 10 th percentile of the baby weights. g)find the first quartile of the baby weights.
The US Air Force requires pilots to have heights between 64 and 77 inches. Women s heights are normally distributed with a mean of 63.8 inches and a standard deviation of 2.6 inches. Men s heights are normally distributed with a mean of 69.5 inches and a standard deviation of 2.4 inches. If US Air Force requirements are changed to exclude only the tallest 3% of men and the shortest 3% of women, what would be the new height requirements? Do You Know How to convert values from a normal distribution to z-scores? How to find areas under a normal curve? How to find the value from a normal distribution corresponding to a given proportion? 7.3 Sampling Distributions and the Central Limit Theorem Introduction In practice, statistical studies involve sampling several, perhaps many, individuals. We often compute numerical summaries of samples, and the most commonly used summary is the sample mean x. If several samples are drawn from a population, they are likely to have different values for x. Because of the value of x varies each time a sample is drawn, x is a random variable, and it has a probability distribution. The probability distribution of x is called the of x. of a Sampling Distribution - I m going to do quite a bit more than the book here
Assume we have a population that is just the numbers {1,3,8}. Let s take random samples from this population of size 2 with replacement. Here are all of the possible samples, and then some sample statistics for each sample. sample Mean Median Range Sample Standard deviation Sample Variance 1,1 1 1 0 0 0 1 1,3 2 2 2 1.4 2 1 1,8 4.5 4.5 7 4.95 24.5 ½ 3,1 2 2 2 1.4 2 1 3,3 3 3 0 0 0 1 3,8 5.5 5.5 5 3.54 12.5 ½ 8,1 4.5 4.5 7 4.95 24.5 ½ 8,3 5.5 5.5 5 3.54 12.5 ½ 8,8 8 8 0 0 0 0 The population mean for this population: µ = Proportion of odds The population median for this population = The population range for this population = The population standard deviation for this population = The population variance for this population: σ 2 = The population proportion of odds for this population = Now calculate the mean of the following: a)mean of the sample means b)mean of the sample medians c)mean of the sample ranges d)mean of the sample standard deviations
e)mean of the sample variances f)mean of the sample proportion of odds: Note some of the similarities between the population parameters and the means of the sample statistics. This is significant - we say a statistic targets a parameter if the mean of the statistic of all samples of the same size is equal to the population parameter. Certain statistics always target parameters (which ones?) Others don t (which ones?) In the case of standard deviation, the sample standard deviation is still the best estimate for population standard deviation, but the other two we will start ignoring The sample standard deviation, and the sample mean, variance, and proportion, will be called estimators or point estimates of their corresponding population parameters. Now, one more time The column labeled mean contains the values of the sample mean for each of the 9 possible samples. The mean of the sampling distribution is the average of these 9 values. The variance of the sampling distribution is the population variances of the 9 sample means, and the same for the rest. The mean is equal to. The variance is equal to. Comparing the mean and variance of the sampling distribution to the population mean and variance, we see that the mean of the sampling distribution μ x = and the variance of the sampling distribution, which is not so obvious, 2 σ x =
Finally, the standard deviation of the sampling distribution σ x = These relationships hold in general. Note that the standard deviation σ x is sometimes called the of the mean. Summary Let x be the mean of a simple random sample of size n, drawn from a population with mean µ and standard deviation σ. The mean of the sampling distribution is The standard deviation of the sampling distribution is, which is sometimes called the of the mean. On the IQ test, the population mean score is 100 with a population standard deviation of 15. A simple random sample of 100 IQ scores were taken. Let x be the mean IQ score of the sample. Find the mean μ x and the standard deviation σ x of x. Solution The mean of x is μ x = = The sample size is n =. Therefore, the standard deviation of x is σ x = = = As it turns out, as sample sizes get larger and larger, the sampling distribution of the sample means is approximately normal. This is called The Central Limit Theorem Let x be the mean of a large (n > 30) simple random sample from a population with mean µ and standard deviation σ.
Then x has an approximately normal distribution, with mean and standard deviation. Procedure for Computing Probabilities with the Central Limit Theorem Step 1: Step 2: Step 3: Step 4: Men s weights are normally distributed with a mean of 172 pounds and a standard deviation of 29 pounds. a)if one man is randomly selected, find the probability that his weight is greater than 180 pounds b)if 100 men are randomly selected, find the probability that they have a mean weight that is greater than 180 pounds.
Women have pulse rates that are normally distributed with a mean of 77.5 beats per minute and a standard deviation of 11.6 beats per minute. a)find the first percentile and 99 th percentile for women s heart rates. b)find the probability that 35 randomly selected women have a mean pulse rate between 70 and 85 beats per minute. c)the probability in part b) should be rather high. Does that mean that if a single woman has a heart rate that is NOT between 70 and 85 beats per minute, that would be unusual? The contents of 40 40 ounce bottles of beverage are measured to see what volume of liquid is in each. This sample has a mean of 40.5 ounces and a standard deviation of 0.8 ounces. Assuming that for all 40 ounce bottles of beverage, the population mean volume is 40.0 ounces and the population standard deviation of volume is 0.8 ounces, find the probability that a sample of 40 bottles has a mean of 40.5 ounces or more. Interpret this result as usual or unusual -> talk hypothesis testing
Do You Know How to construct the sampling distribution of a sample mean? How to find the mean of a sampling distribution of x? How to find the variance and standard deviation of a sampling distribution of x? The Central Limit Theorem? How to use the Central Limit Theorem to compute probabilities involving sample means? 7.4 The Central Limit Theorem for Proportions Definition In a simple random sample of n individuals, let x be the number in the sample who have the characteristic. The sample proportion is pˆ Notation: The population proportion is denoted by The sample proportion is denoted by If several samples are drawn from a population, they are likely to have different values for pˆ. Because the value of pˆ varies each time a sample is drawn, pˆ is a random variable, and it has a probability distribution. The probability distribution of pˆ is called the of pˆ. Going back to of a Sampling Distribution - I m going to do quite a bit more than the book here Assume we have a population that is just the numbers {1,3,8}. Let s take random samples from this population of size 2 with replacement. Here are all of the possible samples, and then some sample statistics for each sample. sample Mean Median Range Standard deviation Variance 1,1 1 1 0 0 0 1 1,3 2 2 2 1.4 2 1 1,8 4.5 4.5 7 4.95 24.5 ½ 3,1 2 2 2 1.4 2 1 3,3 3 3 0 0 0 1 3,8 5.5 5.5 5 3.54 12.5 ½ 8,1 4.5 4.5 7 4.95 24.5 ½ Proportion of odds = pˆ
8,3 5.5 5.5 5 3.54 12.5 ½ 8,8 8 8 0 0 0 0 The mean of the sampling distribution of the proportion of odds is the average of these 9 values of pˆ. The mean is denoted. The mean for this sampling distribution is. The variance of the sampling distribution is the population variance of these 9 values, which can be computed by the method presented in Section 3.2. The variance is denoted. The variance for this sampling distribution is. The standard deviation is the square root of the variance. The standard deviation is denoted. The standard deviation for this sampling distribution is. The values of pˆ and are related to the values of the population proportion p = 2/3 and the 2ˆp sample size n = 3. The relationship for the variance is less obvious than the mean. Specifically, pˆ = = 2ˆp These relationships hold in general. Summary Let pˆ be the sample proportion of a simple random sample of size n, drawn from a population with population proportion p. The mean and standard deviation of the sampling distribution of pˆ are pˆ = pˆ = A Pew Research report indicated that in 2008, 48% of teenagers aged 12-17 played video games on their cell phones. A random sample of 150 teenagers is drawn.
Solution The population proportion is p = and the sample size is n =. Therefore, pˆ = pˆ = The Central Limit Theorem for Proportions Let p be the sample proportion for a sample size of n and population proportion p. If np and n(1 p) then the distribution of p is approximately normal (we will demonstrate this in the next section), with mean and standard deviation pˆ = and pˆ = Procedure for Computing Probabilities with the Central Limit Theorem To compute probabilities involving a sample proportion p, use the following procedure: Step 1: Step 2: Step 3: Step 4: The National Center for Educational Statistics reported that 75% of freshmen entering public high schools in the US in 2005 graduated with their class in 2009. A simple random sample of 135 freshmen is chosen. a)find the probability that less than 80% of freshmen in the sample graduated.
b)find the probability that the sample proportion of students who graduated is between 0.65 and 0.80. c)find the probability that more than 65% of freshmen in the sample graduated. d)would it be unusual if the sample proportion of students who graduated was more than 0.85? Justify. The Bureau of Labor Statistics reported in 2010 that 5% of employed adults in the US held multiple jobs. A random sample of 75 employed adults is chosen. a)would it be appropriate to use the normal approximation to find the probability that less than 6.5% of the individuals in the sample hold multiple jobs? b)a new sample of 350 employed adults is chosen. Find the probability that more than 6% of the individuals in the sample of 350 hold multiple jobs. c) Find the probability that the proportion of individuals in the sample of 350 who hold multiple jobs is between 0.05 and 0.10.
d)would it be unusual if less than 4% of the individuals in the sample of 350 held multiple jobs? Justify. Do You Know The notation for sample and population proportions? How to construct the sampling distribution for a sample proportion? How to find the mean of the sampling distribution of ˆp? How to find the variance and standard deviation of the sampling distribution of ˆp? How to use the Central Limit Theorem to compute probabilities for sample proportions? 7.5 The Normal Approximation to the Binomial Distribution : Imagine that we flip a coin 10 times. Find the probability of getting: a)0 heads b)1 head c)2 heads d)3 heads e)4 heads f)5 heads g)6 heads h)7 heads i)8 heads j)9 heads k)10 heads
l)draw a graph with successes on the x axis and probability on the y-axis. What s the graph look like? Conditions where we can use the normal approximation to the binomial distribution: 7.6 In an Nutshell: A sample is considered to be approximately normal if: 1) 2) 3)