6.1 Graphs of Normal Probability Distributions. Normal Curve aka Probability Density Function

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1 Normal Distributions (Page 1 of 23) 6.1 Graphs of Normal Probability Distributions Normal Curve aka Probability Density Function Normal Probability Distribution TP TP µ! " µ µ +! x x-axis Important Properties of a Normal Curve 1. The curve is bell-shaped with the highest point over the mean µ. 2. It is symmetric about the vertical line through µ. 3. The curve approaches the horizontal axis but never crosses or touches it. 4. The transition points (TP) are where the graph changes from cupping upward to cupping downward (or visa versa). The transition points occur at x = µ +! and x = µ! ". 5. The total area under the curve is 1. Guided Exercise 2 a. Which point (A, B or C) corresponds to µ +!? b. Which point (A, B or C) corresponds to µ! 2"? A B C x c. What is the mean and standard deviation of the distribution?

2 Normal Distributions (Page 2 of 23) Example A The mean affects the location of the curve and the standard deviation affects the shape (spread) of the curve. Same Standard Deviation & Different Means A B Which curve has the larger mean? Same Mean & Different Standard Deviations C Which curve has the larger standard deviation? D

3 Normal Distributions (Page 3 of 23) Guided Exercise 1 Determine whether each curve is normal or not. If it is not, then state why. Empirical Rule for a Normal Distribution a. Approximately 68.2% of the data will lie within 1 standard deviation of the mean. b. Approximately 95.4% of the data will lie within 2 standard deviations of the mean. c. Approximately 99.7% of the data will lie within 3 standard deviations of the mean. d. The area under the curve represents the probability and the total area under the curve is 1. Empirical Rule 68.2% 0.15% 2.15% µ! 3" 13.6% µ! 2" µ! " 13.6% µ µ +! 2.15% µ + 2! µ + 3! 0.15% x-axis

4 Normal Distributions (Page 4 of 23) Example 1 The playing life of a Sunshine radio is normally distributed with a mean of 600 hours and a standard deviation of 100 hours. a. Sketch a normal curve showing the distribution of the playing life of the Sunshine radio. Scale and label the axis; include the transition points. Use the empirical rule to find the probability that a randomly selected radio will last b. between 600 and 700 hours? c. between 400 and 500 hours d. greater than 700 hours?

5 Normal Distributions (Page 5 of 23) Guided Exercise 4 The annual wheat yield per acre on a farm is normally distributed with a mean of 35 bushels and a standard deviation of 8 bushels. a. Sketch a normal curve and shadein the area that represents the probability that an acre will yield between 19 and 35 bushels. b. Is the shaded area the same as the area between µ! 2" and µ? Use the empirical rule to find the probability that the yield will be between 19 and 35 bushels per acre? Control Charts A control chart for a random variable x that is approximately normally distributed is a plot of observed x values in time sequence order. The construction is as follows: 1. Find the mean µ and standard deviation! of the x distribution in one of two ways: (i) Use past data from a period during which the process was in control or (ii) Use a specified target value of µ and!. 2. Create a graph where the vertical axis represents the x values and the horizontal axis represents time. 3. Draw a solid horizontal line at height µ and horizontal dashed control-limit lines at µ ± 2! and µ ± 3!. 4. Plot the variable x on the graph in time sequence order.

6 Normal Distributions (Page 6 of 23) Example 2 Susan is director of personnel at Antlers Lodge and hires many college students every summer. One of the biggest activities for the lodge staff is to make the rooms ready for the next guest. Although the rooms are supposed to be ready by 3:30 pm, there are always some rooms not finished because of high turn over. Every 15 days Susan has a control chart of the number of rooms not made up by 3:30 pm each day. From past experience, Susan knows that the distribution of rooms not made up by 3:30 pm is approximately normal, with µ = 19.3 rooms and! = 4.7 rooms. For the past 15 days the staff has reported the number of rooms not made up by 3:30 pm as: Day x a. Make a control chart for these data. Completely annotate. Control Chart for the Number of Rooms Not Made Up by 3:30 PM µ + 3! µ + 2! µ µ! 2" µ! 3" b. Is the housekeeping process out of control? Explain.

7 Normal Distributions (Page 7 of 23) Out-of-Control Warning Signals 1. Out-of-Control Signal I: One point falls beyond the 3! level. The probability that this is a false alarm is µ + 3! µ + 2! µ µ + 2! µ + 3! 2. Out-of-Control Signal II: A run of 9 consecutive points on one side of the center line. The probability that this is a false alarm is 2!(0.5) 9 = µ + 3! µ + 2! µ µ + 2! µ + 3! 3. Out-of-Control Signal III: At least 2 of 3 consecutive points lie beyond the 2! level on the same side of the center line. The probability that this is a false alarm is µ + 3! µ + 2! µ µ + 2! µ + 3!

8 Normal Distributions (Page 8 of 23) Example 3 Yellowstone Park Medical Services (YPMS) provides emergency medical care for park visitors. History has shown that the during the summer the mean number of visitors treated each day is 21.7 with a standard deviation of 4.2. For a 10-day summer period, the following numbers of people were treated: Day Number Treated a. Make a control chart and plot the data on the chart. µ + 3! µ + 2! µ µ! 2" µ! 3" b. Do the data indicate the number of visitors treated by YPMS is in control or out-of-control? Explain your answer in terms of the three out-ofcontrol signal types. c. If you were the park superintendent, do you think YPMS might need some extra help? Explain.

9 Normal Distributions (Page 9 of 23) 6.2 Standard Units and The Standard Normal Distribution Suppose Tina and Jack are in two different sections of the same course and they recently took midterms. Tina s class average was 64 and she got a 74. Jack s class average 72 and he got an 82. Who did better? z-score or Standardized Score To standardize test scores we use the z-value or (z-score). The z- value, or z-score, or standardized score is the number of standard deviations a data value lies away from the mean. The z-score can be positive, negative, or zero depending on whether the data value is above the mean, below the mean, or at the mean, respectively. For any x-value in a normal distribution the standardized score, z- value, or z-score is given by: z! score = x! µ " Note(s): 1. If x = µ, then z = 0 2. If x > µ, then z > 0 3. If x < µ, then z < 0 z-score negative x < µ µ z-score positive x > µ x Fact: Unless otherwise stated, for now on, the average will mean the arithmetic mean.

10 Normal Distributions (Page 10 of 23) Example 3 Suppose Tina and Jack are in two different sections of the same course and they recently took midterms. The distribution of scores in both classes is normal. Tina s class average was 64 with a standard deviation of 3; she earned a 74. Jack s class average was 72 with a standard deviation of 5; he earned an 82. a. What was Tina s z-score? Draw a distribution for Tina s class showing Tina s score within the distribution of scores for her class. b. What was Jack s z-score? Draw a distribution for Jack s class showing Jack s score within the distribution scores for his class. c. Who did better? Why?

11 Normal Distributions (Page 11 of 23) Example 4 A pizza parlor chain claims a large pizza has 8 oz of cheese with a standard deviation of 0.5 oz. An inspector ordered a pizza and found it only had 6.9 oz of cheese. Franchisee s can lose their store if they make pizzas with 3 standard deviations (or more) of cheese below the mean. Assume the distribution of weights is normally distributed. a. Graph the x-distribution. Label and scale the axis. b. Find the z-score for x = 6.9 oz of cheese. c. Is the franchise in danger of losing its store? Why? d. Find the minimum amount of cheese a franchise can put on a large pizza so it is not in danger of losing its store.

12 Normal Distributions (Page 12 of 23) Guided Exercise 6 The times it takes a student to get to class from home is normally distributed with a mean of 17 minutes and a standard deviation of 3 minutes. a. One day it took 21 minutes to get to class. How many standard deviations from the mean is that? Explain the sign. b. Another day it took 12 minutes to get to class. How many standard deviations from the mean is that? Explain the sign. c. On a third day it took 17 minutes to get to class. How many standard deviations from the mean is that? Explain the sign. z-score Formulae z = x! µ " or x = z! + µ

13 Normal Distributions (Page 13 of 23) Guided Exercise 7 Sam s z-score on an exam is 1.3. If the distribution of scores is normally distributed with a mean of 480 and a standard deviation of 70, what is Sam s raw score. Draw the distribution. Standard Normal Distribution / Curve The standard normal distribution is a normal distribution with mean µ = 0 and standard deviation! = 1. The standard normal curve is the graph of the standard normal distribution. The Standard Normal Curve µ = 0,! = 1 TP TP z

14 Normal Distributions (Page 14 of 23) The Normal Cumulative Density Function The area under the normal curve with mean µ and standard deviation! on the interval from a to b represents the probability that a randomly chosen value for x lies between a and b. It is given by the normal cumulative x a b density function (normalcdf): Access the DISTR menu (2nd > VARS) and Area = P(a < x < b) select 2:normalcdf(lower bound, = normalcdf (a,b,[µ,!]) upper bound, [µ,! ]). Example 6 Find the probability that z is between -1 and 1.!3!2! z Area = P(!1 < z < 1) = normalcdf (,,, ) = To show all work and receive full credit (a) (20-50% of the credit) Sketch the normal curve and shade in the area in question. Label and scale the axis. Put dots at the transition points and tick marks on the axis to 3 standard deviations on each side of the mean. (b) (50-70% of the credit) Compute the area under the curve, hence the probability, showing the probability notation, the TI- 83 function accessed along with its inputs and output. Round probabilities to 4 decimal places.

15 Normal Distributions (Page 15 of 23) Example 7 Find the probability that z is 1. between -3 and 3 2. greater than 1!3!2! z!3!2! z 3. between 0 and greater than 2.53!3!2! z!3!2! z 5. less than between -2 and 2!3!2! z!3!2! z

16 Normal Distributions (Page 16 of 23) 6.3 Area Under Any Normal Curve TI-84: The area under any normal curve between the values a and b is given by Area = P(a < x < b) = normalcdf (a,b,µ,! ) a b x To show all work and receive full credit (a) (20-50% of the credit) Sketch the normal curve and shade in the area in question. Label and scale the axis. Put dots at the transition points and tick marks on the axis to 3 standard deviations on each side of the mean. (b) (50-70% of the credit) Compute the area under the curve, hence the probability, showing the probability notation, the TI- 83 function accessed along with its inputs and output. State probabilities to the nearest ten-thousandth (4 decimal places). Example 7 Given that x has a normal distribution with a mean of 3 and standard deviation of 0.5, find the probability that an x selected at random will be between 2.1 and 3.7. Show your work and include a sketch of the normal curve relevant to this application.

17 Normal Distributions (Page 17 of 23) Example 8 Let x have a normal distribution with a mean of 10 and a standard deviation of 2. Find the probability that an x selected at random from the distribution is between 11 and 14. Show your work and include a sketch of the normal curve representing the probability. Example A A factory has a machine that puts corn flakes in boxes that are advertised as 20 ounces each. If the distribution of weights is normal with µ = 20 and! = 1.5, what is the probability that the weight of a randomly selected box of corn flakes will be between 19 and 21 oz? Show your work and include a sketch of the normal curve representing the probability. Guided Exercise 10 If the life of a Sunshine Stereo is normally distributed with a mean of 2.3 years and a standard deviation of 0.4 years, what is the probability that a randomly selected stereo will break down during the warranty period of 2 years? Show your work and include a sketch of the normal curve representing the probability.

18 Normal Distributions (Page 18 of 23) The Inverse-Normal Function: DISTR / 3:invNorm( P(x < a) a x a = invnorm(area to the left of a, µ,! ) Function Access the DISTR menu (2nd > VARS) and select 3:invNorm(area to the left of a, [ µ,! ]). Input Area to the left of a = P(x < a) Mean µ (default value is 0) Standard Deviation! (default value is 1) Output The value of a on the x-axis of the normal curve. Guided Exercise 11 Find the value of a on the z-axis so that 3% of the area under the standard normal curve lies to the left of a. Round to two decimal places.!3!2! z

19 Normal Distributions (Page 19 of 23) Example B a. Draw a standard normal curve. Then find the value of a > 0 so that 32% of curve lies between 0 and a.!3!2! z b. Draw a standard normal curve. Then find the value(s) of a on the z-axis so that 94% of the curve lies between -a and a.!3!2! z c. Draw a normal curve with a mean of 90 and a standard deviation of 7. Then find the values of b so that 41% of the curve lies between the mean and b. d. Draw a normal curve with a mean of 45 and a standard deviation of 5. Then find the value of b so that 88% of the curve lies between 45 - b and 45 + b.

20 Normal Distributions (Page 20 of 23) Example C Suppose a distribution is normal with a mean of 44 and a standard deviation of 6. Find the value of a (to the nearest hundredth) on the x-axis so that a. 66% of the data values lie below a. b. 15% of the data values lie above a. Example 9 Magic Video Games Inc. sells expensive computer games and wants to advertise an impressive, full-refund warranty period. It has found that the mean life for its computer games is 30 months with a standard deviation of 4 months. If the life spans of the computer games are normally distributed, how long of a warranty period (to the nearest month) can be offered so that the company will not have to refund the price of more than 7% of the computer games?

21 Normal Distributions (Page 21 of 23) Exercise 39 Suppose you eat lunch at a restaurant that does not take reservations. Let x represent the mean time waiting to be seated. It is known that the mean waiting time is 18 minutes with a standard deviation of 4 minutes, and the x distribution is normal. What is the probability that the waiting time will exceed 20 minutes given that it has exceeded 15 minutes? Let event A = x > 20 minutes event B = x > 15 minutes Answer by completing the following: a. In terms of events A and B, what is it we want to compute? b. Is the event B and A = event x > 20 (i.e. event A)? c. Show that P(B and A) = P(B)! P(A, given B) is the same as P(x > 20) = P(x > 15)! P(A, given B) d. Compute P(A), P(B) and P(A, given B)

22 Normal Distributions (Page 22 of 23) 6.4 Approximate a Binomial Distribution with a Normal Distribution Fact If np > 5 and nq > 5 in a binomial distribution, then the sample size n is large enough so that the binomial random variable r has a distribution that is approximately normal. The mean and standard deviation of the normal distribution are estimated by µ = np and! = npq. Furthermore, as the sample size gets larger the approximation gets better.

23 Normal Distributions (Page 23 of 23) Approximating a Binomial Distribution Using a Normal Distribution If np > 5 and nq > 5, then the binomial random variable r has a distribution that is approximately normal. The mean and standard deviation of the normal distribution are estimated by µ = np and! = npq. Example 12 The owner of a new apartment building needs to have 25 new water heaters installed. Assume the probability that a water heater will last 10 years is (a) What is the probability that 8 or more will last at least 10 years. (b) Can the binomial probability distribution be approximated by a normal distribution? Why / Why not? (c) Estimate part (a) with a normal distribution. Take note of the continuity correction. That is, remember to subtract 0.5 from the left endpoint of the interval and add 0.5 to the right endpoint.