Normal Distribution: Calculations of Probabilities

Transcription

1 OpenStax-CNX module: m Normal Distribution: Calculations of Probabilities Irene Mary Duranczyk Suzanne Loch Janet Stottlemyer Based on Normal Distribution: Calculations of Probabilities by Susan Dean Barbara Illowsky, Ph.D. This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 3.0 Probabilities are calculated by using technology and the normal distribution table in the appendix of the text. Example 1 If the area to the left is , then the area to the right is = Example 2 The nal exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of. Problem 1 Find the probability that a randomly selected student scored more than 6 on the exam. Start by dening the variable and describing the population Let X = a score on the nal exam. X N (63, ), where µ = 63 and = Next calculate the z-score, z = x µ = 6 63 = 2 = 0.40 Then draw the normal distribution curve labeling the x-axis with z-scores. Remember that the peak of the distribution will be at zero. Graph where a z-score of will be found. We want the probability of being more than this z=score. P (x > 6). P (z > +0.40) Version 1.2: Aug 21, :23 pm

2 OpenStax-CNX module: m Using the normal distribution table locate the z-score you calculated. The ones and tenths place is found along the side of the table and the hundredths place is located at the top of the table. Our z-score is so we will use the positive z-score table and nd 0.4 on the side and 0.00 on the top. Our probability is found where this row and column intersect. Remember that we are looking for the probability of more than +0.40, P(x > 6) = P(z > +0.40). Our table tells us the probability of being less than our z-score so we need to subtract the probability on the chart from one. Problem 2 Find the probability that a randomly selected student scored less than Using the same population mean and standard deviation in problem 1, let X = score on the nal exam. X N(63, ), where µ = 63 and =. Next calculate the z-score, z = x µ = = 13.3 = 2.66 Then draw the normal distribution curve labeling the x-axis with z-scores. Remember that the peak of the distribution will be at zero. Graph where a z-score of will be found. We want

3 OpenStax-CNX module: m the probability of being less than this z=score, P(x < 76.3) = P(z < +2.66), so we will shade the graph to the left. Figure 1 Our table tells us the probability of being less than our z-score the probability we see on the table is the answer to our question; P (x < 76.3) = P (z < +2.66) = or99.61%. The probability that one student scores less than 76.3 is approximately.9961 (or 99.61%). Problem 3 Find the 90th percentile (that is, nd the score k that has 90 % of the scores below k and 10% of the scores above k). We are using the same population as in the previous problems, X N(63, ). Start by using the standard normal table and nding which z-score gives a probability closest to 90% or To do this look at the probabilities in the interior of the table, is the closest.090. The z-score for this probability is 1.28.

4 OpenStax-CNX module: m Next draw the graph and shading the area that corresponds to the 90 th percentile. Find the z-score 1.28 on graph and shade the area to the right. Using your z-score, mean and standard deviation put the values you know into the z-score formula. z = k µ Use algebra to solve for k, the score closest to 90% 1.28 = k 63 The score closest to 90% is Example 3 A computer is used for oce work at home, research, communication, personal nances, education, entertainment, social networking and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is 2 hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour. Problem 1 Find the probability that a household personal computer is used between 1.8 and 2.7 hours per day. Let X = the amount of time (in hours) a household personal computer is used for entertainment. x N (2, 0.) where µ = 2 and = 0.. Find P (1.8 < x < 2.7). We want to know the probability of being between 1.8 and 2.7 hours. Start by calculating the z-score for each of these values. To calculate the z-score, z = x µ = = 0.2 x µ 0. = 0.40 z = = =.7 0. = 1. Draw the standard normal distribution curve labeling the x-axis with z-scores. Graph where a z-score of and +1.0 will be found. We want the probability of being between these z-scores, P(1.8<x <2.7) = P(-0.40<z <+1.0), so we will shade between these two z-scores.

5 OpenStax-CNX module: m46212 Figure 2 Now look up the two z-scores on the Z-score table to nd their probabilities. P(z<-0.40) = and P(z<1.0) = To nd the probability of being between two values subtract the two probabilities, the larger probability take away the smaller probability, = P ( 0.40 < z < +1.0) = The probability that a household personal computer is used between 1.8 and 2.7 hours per day for entertainment is Problem 2 Find the maximum number of hours per day that the bottom quartile of households use a personal computer for entertainment. To nd the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, nd the 2th percentile, k, where P (x < k) = 0.2.

6 OpenStax-CNX module: m Find the z-score that has a probability closest to 2% or.2. The P(z<-0.67)= Using the same population as in the previous problem, X N(2, 0.) put what you know into the z-score formula and solve for k. z = k µ Use algebra to solve for k, the score closest to 2% 0.67 = k 2 0. The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.