Math 14, Homework 6.2 p. 337 # 3, 4, 9, 10, 15, 18, 19, 21, 22 Name

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1 Name 3. Population in U.S. Jails The average daily jail population in the United States is 706,242. If the distribution is normal and the standard deviation is 52,145, find the probability that on a randomly selected day, the jail population is (a) greater than 750,000. X L 750,000 52, µ 706, P (0 < z < 0.84) Find X µ µ 3 µ 2 µ µ µ+ µ+2 µ+3 L P(X > 750,000) P(z > 0.84 ) % (b) between 600,000 and 700,000. X L 600,000 X U 700,000 µ 706,242 52,145 label the z-axis with, z U, shade the appropriate X L µ z U P ( 2.04 < z < 0) P ( 0.12 < z < 0) Find µ 3 µ 2 µ µ µ+ µ+2 µ+3 and z U X µ U P(600,000 < X < 700,000) P( 2.04 < z < 0.12) %

2 4. SAT Scores The national average SAT score (for Verbal and Math) is Assume a normal distribution with 92. (a) What is the 90 th percentile score?, label the z-axis with z A, A µ 1028 or statistics calculator to find the z A -value: z 1.28 µ 3 µ 2 µ µ µ+ µ+2 µ Use the formula, z A X µ to solve for the desired score X X The 90 th percentile score is (b) What is the probability that a randomly selected score exceeds 1200? X L 1200 µ Find X µ L µ 3 µ 2 µ µ µ+ µ+2 µ P(X > 1,200) P(z > 1.87) %

3 9. Miles Driven Annually The mean number of miles driven per vehicle annually in the United States is 12,494. Assume the distribution is normal with a standard deviation of For a randomly chosen vehicle, what is the probability that the vehicle was (a) driven more than 15,000 miles? X L 15,000 µ 12, label the z-axis with, P (0 < z < 1.94) Find X µ µ 3 µ 2 µ µ µ+ µ+2 µ+3 L ,494 15, P(X > 15,000) P(z > 1.94) % (b) driven less than 8000 miles? X U 8,000 µ 12,494 1, P (0 < z < 1.94) µ 3 µ 2 µ µ µ+ µ+2 µ+3 Step 4 Find z U X µ U z U , P(X < 8,000) P(z < 3.48) % (c) Would you be comfortable with buying a vehicle for which you had been told it had been driven less than 6000 miles in the last year? No. Only a fraction of 1% of all used cars are driven less than 8000 miles per year. I would need to find out why it had been driven so little in the previous year. 12,494

4 10. Commute Time to Work The average commute to work (one way) is 25 minutes according to the 2005 American Community Survey. If we assume that commuting times are normally distributed and that the standard deviation is 6.1 minutes, what is the probability that a randomly selected commuter spends (a) more than 30 minutes commuting one way? X L 30 µ label the z-axis with, P (0 < z < 0.82) Step 4 Find X µ L µ 3 µ 2 µ µ µ+ µ+2 µ P(X > 30) P(z > 0.82) % (b) less than 18 minutes? X U 18 µ 25 z U P (0 < z < 1.15) Find z U X µ µ 3 µ 2 µ µ µ+ µ+2 µ+3 U P(X < 18) P(z < 1.15) %

5 15. Jobs for Registered Nurses The average annual number of jobs available for registered nurses is 103,900. If we assume a normal distribution with a standard deviation of (a) What is the probability that more than 100,000 jobs are available for RNs? X L 100,000 µ 103,900 8,040 label the z-axis with, P (0 < z < 0.49) Find X µ L µ 3 µ 2 µ µ µ+ µ+2 µ z P(X > 100,000) P(z > 0.49) % (b) What is the probability that more than 80,000 but less than 95,000 jobs are available for RNs? X L 80,000 X U 95,000 X L µ z U 8040 µ 103,900 8,040 label the z-axis with, z U, shade the appropriate P ( 2.97 < z < 0) P ( 1.11 < z < 0) Find and z U X µ µ 3 µ 2 µ µ µ+ µ+2 µ+3 U , , ,900 P(80,000 < X < 95,000) P( 2.97 < z < 1.11) %

6 15. (c) If the probability is that more than X amount of jobs are available, find the value of X. A ,040 µ 103,900 or statistics calculator to find the z A -value: z 0.85, label the z-axis with z A, 103, , µ 3 µ 2 µ µ µ+ µ+2 µ+3 Use the formula, z A X µ to solve for the desired number of jobs X X X 110, ,734 jobs available (has a probability of.1977 of occurring) 18. Itemized Charitable Contributions The average charitable contribution itemized per income tax return in Pennsylvania is $792. Suppose that the distribution of contributions is normal with a standard deviation of $103. Find the limits for the middle 50% of contributions. A µ 792, label the z-axis with, and z U Use the formulas, X µ L and z U X µ U and solve for the desired lower and upper limits X X X or statistics calculator to find the lower and upper z A -values z µ 3 µ 2 µ µ µ+ µ+2 µ+3 The middle 50% of contributions lie between $ and $

7 19. New Home Size A contractor has decided to build homes that will include the middle 80% of the market If the average size of homes built is 1810 square feet, find the maximum and minimum sizes of the homes the contractor should build. Assume that the data is normally distributed and the standard deviation is 92 square feet. A µ 1810 or statistics calculator to find the lower and upper z A -values z 1.28, label the z-axis with, and z U µ 3 µ 2 µ µ µ+ µ+2 µ Use the formulas, X µ L and z U X µ U and solve for the desired minimum and maximum sizes X X The middle 80% of the market has homes between 1692 and 1928 square feet. X Cost of Personal Computers The average price of a personal computer is $949. Assume the computer prices are approximately normally distributed and $100. (a) What is the probability that a randomly selected PC costs more than $1200? X L 1200 µ Find X µ L µ 3 µ 2 µ µ µ+ µ+2 µ P(X > 1,200) P(z > 2.51) %

8 21. (b) The least expensive 10% of personal computers cost less than what amount? A µ 941 or statistics calculator to find the z A -value: z 1.28, label the z-axis with z A, µ 3 µ 2 µ µ µ+ µ+2 µ Use the formula, z A X µ to solve for the desired computer cost X X 821 The least expensive 10% cost less than $ Reading Improvement Program To help students improve their reading skills, a school district decides to implement a reading program. It is to be administered to the bottom 5% of the students in the district, based on the scores on a reading achievement exam. If the average score for the students in the district is 122.6, find the cutoff score that will make a student eligible for the program. The standard deviation is 18. Assume the data is normally distributed. A µ 941 or statistics calculator to find the z A -value: , label the z-axis with z A, µ 3 µ 2 µ µ µ+ µ+2 µ Use the formula, z A X µ to solve for the desired cutoff score X X The cutoff score of the lowest 5% of students is 93. z 1.64

9 Area Standard Normal Distribution Table z