Chapter 6 - Normal Probability Distributions

Normal Distribution

If a continuous random variable is said to be "normally distributed", this means: the probability density curve is bell-shaped and symmetric This shape is called the normal probability distribution or the normal curve.

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 8 minutes. Find the probability that a randomly selected passenger has a waiting time less than 0.75 minutes.

P(less than 0.75) = (length of shaded region)*(height of shaded region) = (0.75-0)*(0.13) = (.75)*(.13) = .098

Z_α Notation

- For the standard normal distribution, a critical value is a z score separating unlikely values from those that are likely to occur. - Notation: The expression z_α (z, with subscript alpha) denotes the z score with an area to its RIGHT equal to α (alpha). - When a decimal value is given as a subscript with a z-score, Z_α this notation indicates that the z-score creates an area to its RIGHT equal to the subscript.

Find the indicated critical value. z_0.08

1- 0.08 = 0.92 z_0.08 = 1.41 go to 0.92 on table and find out z score

Find the indicated IQ score. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. (0.4, shaded to the right)

1-0.4 = 0.6 area = 0.6 -> z = 0.25 x = 100 + (.25*15) = 103.8

Properties of a Normal Distribution

1. Bell-shaped and is symmetric about the mean (that means it is centered at μ) 2. As the graph extends farther and farther away from the mean, it approaches the x-axis asymptotically (like an asymptote) meaning the graph gets closer and closer to the x-axis but never touches it. 3. The graph has inflection points where the curve changes from concave down (in the center) to concave up (at both ends). The distance from the center to the inflection points is σ 4. The total area that lies under the curve is one (100%) 5. The area to the right of µ is equal to the area to the left of µ--> both are 1/2 (50%)

Finding Values From Known Areas

1. Don't confuse x-values and probability areas. X-values (or z-scores) are distances along the horizontal scale, Probability areas are regions under the probability curve. 2. Think before you draw. Label your x-values (z-scores) or probability area on the correct side of the mean. 3. Choose the correct (>right, or < left) side of the graph to shade.

Procedure for Determining Whether It Is Reasonable to Assume that Sample Data are From a Normally Distributed Population

1. Histogram: Construct a histogram. Reject normality if the histogram departs dramatically from a bell shape. 2. Outliers: Identify outliers. Reject normality if there are one or more outliers. 3. Normal Quantile Plot: Use technology to generate a normal quantile plot. In StatCrunch: Select GRAPH > QQ Plot

Central Limit Theorem

1. The distribution of sample means, x bar, will be normally distributed with the following predictable characteristics: 2. The mean of the sample means is equal to the population mean 3. The standard deviation of the sample means is smaller than the population std. deviation. -- The standard deviation of the sample means is the population standard deviation divided by the square root of the sample size

Continuous probability density curves must satisfy the following properties:

1. The total area under the curve must equal 1. 2. The entire curve must be ≥ 0. (That is, the curve cannot fall below the x-axis.)

Finding Probability from Known z-Scores

1.Draw a normal curve. Shade the area of interest. 2. Use technology (StatCrunch or other Stat calculator) or Table A-2 (in the "Appendix" of the e-book or back cover of the printed book) to find the corresponding probability area.

Uniform Distribution

A continuous random variable has a uniform distribution if its values are spread evenly over the range of probabilities.

Normal Quantile Plots

A normal quantile plot (or normal probability plot) compares - the original sample data (sorted) on x-axis - expected quantile z-scores on y-axis

The population of current statistics students has ages with mean mu and standard deviation sigma​, with an unknown shape of the population distribution. Samples of statistics students are randomly selected so that there are exactly 46 students in each sample. For each​ sample, the mean age is computed. What does the central limit theorem tell us about the distribution of those mean​ ages?

Because n > ​30, the sampling distribution of the mean ages can be approximated by a normal distribution with mean mu and standard deviation sigma/sqrt(46)

Area = Probability

Because the total area under the density curve is equal to 1: The area under the curve corresponds to the probability assigned to the random variable.

The​ _______ tells us that for a population with any​ distribution, the distribution of the sample means approaches a normal distribution as the sample size increases.

Central Limit Theorem

Important Power of the Central Limit Theorem

Even when the parent population is NOT normally distributed.... ...As the sample size increases, the sampling distribution of sample means approaches a normal distribution. Case 1: If the parent population is KNOWN to be normally distributed, use a sample size: n = any size Case 2: If it is unknown whether or NOT the parent population is normally distributed, use a sample size n >= 30

Summary: Key Points from 6.5

Regardless of the sample size: 1. The mean of the sampling distribution is equal to the mean of the parent population 2. And the standard deviation of the sampling distribution is sigma/sqrt(n) 3. The Central Limit Theorem: The shape of the distribution of sample means becomes approximately normal as the sample size n increases (n =30 or more) even if the parent population is NOT normal

Hypothesis Tests for Normality

The Shapiro-Wilk or Anderson-Darling Normality tests are two of several formal hypothesis tests of normality, each having their own advantages and disadvantages. StatCrunch includes these two Normality Assessments both of which can displays a histogram and normal quantile plot, along with the test statistic and p-value results for the test. STAT> Goodness of Fit> Normality Test ...check the options for QQ plot and Histogram

Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation 15. Find P4​, which is the IQ score separating the bottom 4% from the top 96​%.

The area closest to 0.04 is .0401 z = -1.75 x = 100 + (-1.75*15) = 73.75

Which of the following is NOT a conclusion of the Central Limit​ Theorem?

The distribution of population data must always start as a normal​ distribution, for the distribution of the sample means x overbar to approach a normal distribution.

Which of the following is NOT a requirement for a density​ curve?

The graph is centered around 0.

Which of the following does NOT describe the standard normal​ distribution?

The graph is uniform.

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 50.0 and 55.0 minutes. Find the probability that a given class period runs between 50.5 and 51.0 minutes.

The length of the uniform distribution is the difference between the maximum and minimum values. The length of the uniform distribution is 5. Since the uniform distribution is​ rectangular, has a length of 5.0​, and an area of​ 1, the height of the uniform distribution is .2. P(between 50.5 and 51.0) = (shaded region length)*(shaded region height) = (51.0-50.5)*(0.2) = (0.5)(0.2) = 0.1

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 6 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 3.25 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 3.25 minutes.

The length of the uniform distribution is the difference between the maximum and minimum values. The length of the uniform distribution is: 6 Since the uniform distribution is​ rectangular, has a length of 6​, and an area of​ 1, determine the height of the uniform distribution. The height of the uniform distribution is: .17 P(greater than 3.25) = (length of shaded region)*(height of shaded region) = (6-3.25)(0.17) = (2.75)(0.17) = 0.468

What requirements are necessary for a normal probability distribution to be a standard normal probability​ distribution?

The mean and standard deviation have the values of µ = 0 and σ = 1.

Intro to Probability Curves

The probabilities assigned to a continuous random variable (like time, heights, distances, measurements) can't be listed...so they are usually shown in a graph.

Sampling Distributions

The sampling distribution of the sample mean, x bar is the distribution that is formed when samples of size n are repeatedly taken from a population.

Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

The sketch to the right can be used to illustrate the general rule for finding the area between two z scores. Note that the shaded region B can be found by calculating the difference between the area to the left of z= -1.21 and the area to the left of z=1.19. The area to the left of z=1.19 is 0.8830​, rounding to four decimal places. The area to the left of z=-1.21 is 0.1131​, rounding to four decimal places. Area = .8830-.1131 = .7699

Non-"Standard" Normal

This section presents methods for working with the general normal distributions that are not standard. That is, the mean may be somewhere other than 0 and the std. dev. may be something other than 1.

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

To find the area of the shaded​ region, find the area below z=0.39. Using a standard normal distribution​ table, begin with the z score of 0.39 by locating 0.3 in the left​ column; next find the value in the adjoining row of probabilities that is directly below 0.09.

Is the statement below true or​ false? The distribution of the sample​ mean, x overbar​, will be normally distributed if the sample is obtained from a population that is normally​ distributed, regardless of the sample size.

True

Assume that adults have IQ scores that are normally distributed with a mean of 104 and a standard deviation of 15. Find the third quartile Q3​, which is the IQ score separating the top​ 25% from the others.

Use the standard normal distribution table to find the z score that corresponds to an area of​ 75%, or 0.75. The z score is .67 x = u+(z*o) = 104+(.67*15) = 114.1

Testing Normality

We can conclude that the sample data is taken from a population that is normally distributed, if... the pattern of the points is reasonably close to a straight line and the points do not show some systematic pattern that is not a straight-line pattern.

A ski gondola carries skiers to the top of a mountain. It bears a plaque stating that the maximum capacity is 18 people or 2790 lb. That capacity will be exceeded if 18 people have weights with a mean greater than 2790lb/18 = 155lb. Assume that weights of passengers are normally distributed with a mean of 169 lb and a standard deviation of 38.6 lb. Complete parts a through c below. a. Find the probability that if an individual passenger is randomly​ selected, their weight will be greater than 155 lb. b. If 18 passengers are randomly​ selected, find the probability that their mean weight is greater than 155 lb. c. Does the gondola appear to have the correct weight​ limit? Why or why​ not?

a. z = (155-169)/38.6 = -0.36 area = .3594 1-.3594 = .6406 b. z = (155-169)/38.6/sqrt(18) = -1.54 area = .0618 1-.0618 = .9382 ​No, there is a high probability that the gondola will be overloaded if it is occupied by 18 ​passengers, so it appears that the number of allowed passengers should be reduced.

Women have pulse rates that are normally distributed with a mean of 80.2 beats per minute and a standard deviation of 10.9 beats per minute. Complete parts a through c below. a. Find the percentiles P 1 and P 99 b. A doctor sees exactly 30 patients each day. Find the probability that 30 randomly selected women have a mean pulse rate between 66 and 94 beats per minute. c. If the doctor wants to select pulse rates to be used as cutoff values for determining when further tests should be​ required, which pulse rates are better to​ use: the results from part​ (a) or the pulse rates of 66 and 94 beats per minute from part​ (b)? Why?

a. z 0.01 = -2.33 z 0.99 = 2.33 P 1 = (10.9)(-2.33) + 80.2 = 54.8 P 99 = (10.9)(2.33) + 80.2 = 105.6 b. z = (66-80.2) / (10.9/sqrt(30)) = -7.14 z = (94-80.2) / (10.9/sqrt(30)) = 6.93 The area to the left of z=-7.14 is 0.0000. The area to the left of z=6.93 is 1.0000. 1 - 0 = 1 c. Part​ (a), because the cutoff values should be based on individual patients rather than the mean pulse rate of the sample.

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 307 days or longer. b. If the length of pregnancy is in the lowest 4​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

a. z = (307-266)/15 = 2.73 P(z<2.73) = .9968 0.0032 b. .0401 z = -1.75 x = 266 + (-1.75*15) = 240

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 307 days or longer. b. If the length of pregnancy is in the lowest 3​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

a. z = (307-266)/15 = 2.73 P(z<2.73) = .9968 shaded area = 1-.9968 = 0.0032 b. closest to 0.03 is 0.0301 z = -1.88 x = 266 + (-1.88*15) = 238

Women have head circumferences that are normally distributed with a mean given by mu=22.38 in.​, and a standard deviation given by sigma equals 0.6 in. Complete parts a through c below. a. If a hat company produces​ women's hats so that they fit head circumferences between 22.0 in. and 23.0 ​in., what is the probability that a randomly selected woman will be able to fit into one of these​ hats? b. If the company wants to produce hats to fit all women except for those with the smallest 1.5​% and the largest 1.5​% head​ circumferences, what head circumferences should be​ accommodated? c. If 21 women are randomly​ selected, what is the probability that their mean head circumference is between 22.0in. and 23.0​in.? If this probability is​ high, does it suggest that an order of 21 hats will very likely fit each of 21 randomly selected​ women? Why or why​ not? (Assume that the hat company produces​ women's hats so that they fit head circumferences between 22.0 in. and 23.0 ​in.)

a. z = 22-22.38/0.6 = -.63 z = 23-22.38/0.6 = 1.03 The area to the left of z=-0.63 is .2643. The area to the left of z=1.03 is .8485. .8458 - .2643 = .5842 b. z0 = -2.17 z1 = 2.17 cmin = (0.6)(-2.17) + 22.38 = 21.08 cmax= (0.6)(2.17) + 22.38 = 23.68 c. z = 22-22.38/(0.6/sqrt(21)) = -2.9 z = 23-22.38/(0.6/sqrt(21)) = 4.74 left of z = -2.9 is 0.0019 left of z = 4.74 is 1.0000 1.00000 - 0.0019 = .9981

A boat capsized and sank in a lake. Based on an assumption of a mean weight of 130 ​lb, the boat was rated to carry 70 passengers​ (so the load limit was 9,100 ​lb). After the boat​ sank, the assumed mean weight for similar boats was changed from 130 lb to 170 lb. Complete parts a and b below.

a. 1 b. .7210 Because there is a high probability of​ overloading, the new ratings do not appear to be safe when the boat is loaded with 17 passengers.

Suppose a simple random sample of size n = 15 is obtained from a population with mu = 68 and sigma = 16. ​(a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities regarding the sample​ mean? Assuming the normal model can be​ used, describe the sampling distribution x overbar. ​(b) Assuming the normal model can be​ used, determine ​P(x overbar < 71.7​). ​(c) Assuming the normal model can be​ used, determine ​P(x overbar >= 69.2​).

a. The population must be normally distributed b. normal, with u_x = 68 and o_x = 16/sqrt(15) c. z = (x-bar - u_x) / o_x, find area left of z P(x < 71.7) =

Assume that human body temperatures are normally distributed with a mean of 98.21 degrees F and a standard deviation of 0.63 degrees F. a. A hospital uses 100.6 degrees F as the lowest temperature considered to be a fever. What percentage of normal and healthy persons would be considered to have a​ fever? Does this percentage suggest that a cutoff of 100.6 degrees F is​ appropriate? b. Physicians want to select a minimum temperature for requiring further medical tests. What should that temperature​ be, if we want only​ 5.0% of healthy people to exceed​ it? (Such a result is a false​ positive, meaning that the test result is​ positive, but the subject is not really​ sick.)

a. z = (100.6-98.21)/0.63 = 3.79 0.01% Yes, because there is a small probability that a normal and healthy person would be considered to have a fever. b. The z score with an area of​ 95%, or​ 0.95, to its left is 1.645 x = 98.21 + (1.645*0.63) = 99.25

Assume that the Richter scale magnitudes of earthquakes are normally distributed with a mean of 1.102 and a standard deviation of 0.503. Complete parts a through c below. a. Earthquakes with magnitudes less than 2.000 are considered​ "microearthquakes" that are not felt. What percentage of earthquakes fall into this​ category? b. Earthquakes above 4.0 will cause indoor items to shake. What percentage of earthquakes fall into this​ category? c. Find the 95th percentile. Will all earthquakes above the 95th percentile cause indoor items to​ shake?

a. z = (2.000-1.102)/0.503 = 1.79 P(z<1.79) = .9629 93.33% b. z = (4.0-1.102)/0.503 = 5.76 P(z<5.76) = 1 0% c. z = 1.645 x = 1.102 + (1.645*.503) = 1.929 ​ No, because not all earthquakes above the 95th percentile have magnitudes above 4.0.

​Men's heights are normally distributed with mean 68.5 in and standard deviation of 2.8 in. ​Women's heights are normally distributed with mean 63.1 in and standard deviation of 2.5 in. The standard doorway height is 80 in. a. What percentage of men are too tall to fit through a standard doorway without​ bending, and what percentage of women are too tall to fit through a standard doorway without​ bending? b. If a statistician designs a house so that all of the doorways have heights that are sufficient for all men except the tallest​ 5%, what doorway height would be​ used?

a. z = (80-68.5)/2.8 = 4.11 P(z<4.11) = 0.9999 1-0.9999 = 0.0001 0.01% b. area left of z-score is .9500 z = 1.645 x = 68.5 + (1.645*2.8) = 73.1

Assume that​ women's heights are normally distributed with a mean given by u=62.5 in​, and a standard deviation given by sigma equals 2.3 in. ​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63 in. ​(b) If 32 women are randomly​ selected, find the probability that they have a mean height less than 63 in.

a. z = 63-62.5/2.5 = .22 area left of z=.22 -> .5871 b. u = 62.5 o = 2.3/sqrt(32) = .406586 z = 63-62.5 / .406586 = 1.23 probability = .8907

Find the indicated IQ score. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

area of 0.45 -> z = -.13 x = 100 + (-0.13*15) = 98.1

Finding probabilities associated with distributions that are standard normal distributions is equivalent to​ _______.

finding the area of the shaded region representing that probability.

Z-scores

follow the standard normal distribution with μ = 0 and σ = 1

Standard Normal Distribution

is drawn with zero at the mean, and standard deviation of 1; μ = 0 and σ = 1 The total area under the density curve equals 1

The standard deviation of the distribution of sample means is​ _______.

sigma / sqrt(n)

If the parent population is normally distributed

x bar will be normally distributed for sample size, n = any size

If the parent population is NOT NORMALLY DISTRIBUTED

x bar will be normally distributed for sample sizes, n >= 30

Find the area of the shaded region. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. (108->132)

z = (108-100)/15 = .53 z = (132-100)/15 = 2.13 area left of z = 0.53 is .7019 area left of z =2.13 is .9834 .9834-.7019 = 0.2815

Find the area of the shaded region. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. (75-110)

z = (75-100)/15 = -1.67 z = (x-u)/o = (110-100)/15 = .67 x = 75 is z = -1.67 x = 110 is z = 0.67 The area to the left of z = -1.67 is .0475 area left of z = 0.67 is .7486 the area is 0.7486 - 0.0475 = .7011

Assume that adults have IQ scores that are normally distributed with a mean of u equals 100 and a standard deviation sigma equals 15. Find the probability that a randomly selected adult has an IQ between 82 and 118.

z = (82-100)/15 = -1.2 P(z < -1.2) = .1151 z = (118-100)/15 = 1.2 P(z < 1.2) = .8849 Shaded area = .8849 - .1151 = .7698

Find the area of the shaded region. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

z = (x-u)/o = (110-100)/15 = .67 area left of z = 0.67 is .7486

Chocolate chip cookies have a distribution that is approximately normal with a mean of 23.4 chocolate chips per cookie and a standard deviation of 2.7 chocolate chips per cookie. Find P 5 and P 95. How might those values be helpful to the producer of the chocolate chip​ cookies?

z = -1.64 x = 23.4 + (-1.64*2.7) = 19.0 z = 1.64 x = 23.4 + (1.64*2.7) = 27.8 The values can be used to identify cookies with an unusually low or high number of chocolate​ chips, so those numbers can be used to monitor the production process to ensure that the numbers of chocolate chips stays within reasonable limits.