Normal & Standard Normal Distributions Chapter 5 Review Normal & Standard Normal Distributions

Find the area of the indicated region under the standard normal curve Find the area of the indicated region under the standard normal curve. Use the table and show your work. Find the areas to the left of the z-scores using the z-score table. The area to the left of the z-score of 0 is .5000 The area to the left of the z-score of 1.2 is 8849 Subtract the two areas. .8849−.5000=.3849 𝑷 𝟎<𝒛<𝟏.𝟐 =.𝟑𝟖𝟒𝟗

Find the indicated area under the standard normal curve Find the indicated area under the standard normal curve. You may find the area using the table and check it in the calculator using normalcdf. To the left of 𝑧=1.36. Sketch the standard normal curve Using the table, the area to the left of 𝑧=1.36 is .9131 𝑷 𝒛<𝟏.𝟑𝟔 =.𝟗𝟏𝟑𝟏 0.9131

Using the graphing calculator: Set the mode to a float of 4. 2nd Vars (Distr) Normalcdf Lower: −1000 Upper: 1.36 𝜇: 0 𝜎: 1 Paste to the home screen and push Enter The area to the left of 𝑧 = 1.36 is .9131 𝑷 𝒛<𝟏.𝟑𝟔 =.𝟗𝟏𝟑𝟏

Find the indicated area under the standard normal curve Find the indicated area under the standard normal curve. Use the table and show your work for credit. To the right of 𝑧=−0.65 Sketch the standard normal curve 1.0000−.2578=.7422 The area at 𝑧=−0.65 is .2578. This is the area to the left of the z-score. To find the area to the right of the z-score, subtract from 1. 1.0000−.2578=.7422 𝑷 𝒛>𝟎.𝟔𝟓 =𝟎.𝟕𝟒𝟐𝟐

Using the graphing calculator to check: 2nd Vars (Distr) Normalcdf Lower: −0.65 Upper: 1000 𝜇: 0 𝜎: 1 Paste to the home screen and push Enter To find the area to the right of the z-score, subtract from 1. .2578 1.0000−.2578=.7422 𝑷 𝒛>𝟎.𝟔𝟓 =𝟎.𝟕𝟒𝟐𝟐

Find the indicated area under the standard normal curve Find the indicated area under the standard normal curve. Use the table and show your work for credit. Between 𝑧=−1.53 and 𝑧=0 Sketch the standard normal curve Find the areas to the left of the two z-scores using the table. The area to the left of 𝑧=−1.53 is .0630 The area to the left of 𝑧=0 is .5000 Subtract the two areas. .5000−.0630=.4370 𝑷 −𝟏.𝟓𝟑<𝒛<𝟎 =𝟎.𝟒𝟑𝟕𝟎

Using the graphing calculator to check: 2nd Vars (Distr) Normalcdf Lower: −1.53 Upper: 0 𝜇: 0 𝜎: 1 Paste to the home screen and push Enter 𝑷 −𝟏.𝟓𝟑<𝒛<𝟎 =𝟎.𝟒𝟑𝟕𝟎

To the left of 𝑧=−1.28 or to the right of 𝑧=1.28 Find the indicated area under the standard normal curve. Use the table and show your work for credit. To the left of 𝑧=−1.28 or to the right of 𝑧=1.28 Find the area to the left of 𝑧=−1.28 .1003 Find the area to the left of 𝑧=1.28 .8997 The area to the right of 𝑧=1.28 is 1.0000−.8997= .1003

Find the area to the left of 𝑧=−1.28 .1003 The area to the right of 𝑧=1.28 is 1.0000−.8997= .1003 What is the clue word? or Which means we’re going to do what? add .1003+.1003=.2006 𝑷 𝒛<−𝟏.𝟐𝟖 𝐨𝐫 𝒛>𝟏.𝟐𝟖 =.𝟐𝟎𝟎𝟔

Find the areas to the left of both z-scores and subtract. Find the probability of z occurring in the indicated region. Show your work for credit. Use the table. The four-digit number in the table is also the probability. Find the areas to the left of both z-scores and subtract. The area to the left of 𝑧=−0.5 is .3085 The area to the left of 𝑧=1.00 is .8413 𝑷 −𝟎.𝟓<𝒛<𝟏.𝟎𝟎 =.𝟖𝟒𝟏𝟑−.𝟑𝟎𝟖𝟓=.𝟓𝟑𝟐𝟖 .

Find the indicated probability using the standard normal distribution Find the indicated probability using the standard normal distribution. Show your work for credit. 𝑃(𝑧<1.45) Use the table. The four-digit number in the table is also the probability. The area to the left of 𝑧=1.45 is .9265 𝑷 𝒛<𝟏.𝟒𝟓 =.𝟗𝟐𝟔𝟓

Find the indicated probability using the standard normal distribution Find the indicated probability using the standard normal distribution. Show your work for credit. 𝑃(𝑧>−0.95) Use the table. The four-digit number in the table is also the probability. The four digit number at 𝑧=−0.95 is .1711 .1711 is the area to the left. We must subtract this from 1. 𝑷 𝒛>−𝟎.𝟗𝟓 =𝟏.𝟎𝟎𝟎𝟎−.𝟏𝟕𝟏𝟏=.𝟖𝟐𝟖𝟗

Find the indicated probability using the standard normal distribution. 𝑃(−1.89<𝑧<0) Sketch the curve. Find the area to the left of the z-scores and subtract. The area to the left of 𝑧=−1.89 is .0294 The area to the left of 𝑧=0 is .5000 The four-digit number represents both area and probability, so subtract the two areas. 𝑷 −𝟏.𝟖𝟗<𝒛<𝟎 =.𝟓𝟎𝟎𝟎−.𝟎𝟐𝟗𝟒=.𝟒𝟕𝟎𝟔

Assume the random variable is normally distributed with a mean 𝜇=86 and standard deviation 𝜎=5. Find the indicated probability. 𝑃(𝑥<80) Here we need to convert the x-value to a z-score in order to find the probability. 𝑧= value−mean standard deviation = 𝑥−𝜇 𝜎 𝑧= 80−86 5 = −6 5 =−1.20 The area to the left of 𝑧=−1.20 is .1151 𝑷 𝒙<𝟖𝟎 =𝑷 𝒛<−𝟏.𝟐𝟎 =.𝟏𝟏𝟓𝟏

Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed. We need to convert the two x-values to z-scores. 𝑧= value−mean standard deviation = 𝑥−𝜇 𝜎 𝑧= 200−186 35.8 = 14 35.8 =0.3911≈0.39

Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed. We need to convert the other x-value to a z-score. 𝑧= value−mean standard deviation = 𝑥−𝜇 𝜎 𝑧= 239−186 35.8 = 53 35.8 =1.4804≈1.48

Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed. Use the table to find the probabilities to the two z-scores and subtract. The area to the left of 𝑧=0.39 is .6517 The area to the left of 𝑧=1.48 is .9306

Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed. 𝑷 𝟐𝟎𝟎<𝒙<𝟐𝟑𝟗 =𝑷 𝟎.𝟑𝟗<𝒛<𝟏.𝟒𝟖 =.𝟗𝟑𝟎𝟔−.𝟔𝟓𝟏𝟕=.𝟐𝟕𝟖𝟗

We have been given x-values, a mean, and a standard deviation We have been given x-values, a mean, and a standard deviation. The heights are normally distributed. So in order to find the probabilities, we need z-scores. And we will need to sketch three curves. Have fun! a) Find the probability that his height is less than 66 inches. 𝑃(𝑥<66 inches) We need to find the z-score and go to the table. 𝑧= value−mean standard deviation = 𝑥−𝜇 𝜎

𝑧= 66−69.6 3.0 = −3.6 3.0 =−1.20 The probability to the left of 𝑧=−1.20 is .1151 𝑷 𝒙<𝟔𝟔 𝐢𝐧𝐜𝐡𝐞𝐬 =𝑷 𝒛<−𝟏.𝟐𝟎 =.𝟏𝟏𝟓𝟏

b) Find the probability that his height is between 66 and 72 inches. We need to find two z-scores and go to the table. 𝑧= value−mean standard deviation = 𝑥−𝜇 𝜎 . We already know one of the z-scores. 𝑧= 66−69.6 3 = −3.6 3 =−1.20

Now find the other z-score. 𝑧= value−mean standard deviation = 𝑥−𝜇 𝜎 . 𝑧= 72−69.6 3.0 = 2.4 3.0 =0.80 The area to the left of 𝑧=−1.20 is .1151 The area to the left of 𝑧=0.80 is .7881

Subtract the two areas (probabilities). ,7881−.1151=.6730 . 𝑷 𝟔𝟔<𝒙<𝟕𝟐 =𝑷 −𝟏.𝟐𝟎<𝒛<𝟎.𝟖𝟎 =.𝟔𝟕𝟑𝟎

The z-score at 72 is 0.80. The probability to the left is .7881 c) Find the probability that his height is more than 72 inches The z-score at 72 is 0.80. The probability to the left is .7881 . The probability to the right is 1−.7881=.2119 𝑷 𝒙>𝟕𝟐 =.𝟐𝟏𝟏𝟗

Missing Homework Assignments Find the z-score that corresponds to the given cumulative area. If the area is not in the table, use the entry closest to the area. 0.7580 Go to the table and look for the area. Match it with the z-score. .7580 can be found in the table. The z-score which matches up with .7580 is 0.70 Missing Homework Assignments

𝑃 20 Find the z-score that corresponds to the given percentile. This is the 20th percentile, which converts to an area of .2000 Go to the table and find the area nearest to .2000 .The area nearest to .2000 is .2005 The z-score that matches with the area .2005 is −0.84 𝒛=−𝟎.𝟖𝟒

Convert 11.9% to a four-digit decimal. .1190 Find the z-score that has 11.9% of the distribution’s area to its left. Convert 11.9% to a four-digit decimal. .1190 Go to the table and find the area closest to .1190 The area nearest to .1190 is exactly .1190 The z-score which matches up to the area .1190 is −1.18 𝒛=−𝟏.𝟏𝟖

Convert 23.7% to a four-digit decimal. .2370 Find the z-score that has 23.7% of the distribution’s area to its right. Convert 23.7% to a four-digit decimal. .2370 This is the area to the right. We need to find the area to the left and find its z-score. 1.0000−.2370= .7630 Find the area nearest to .7630 .7642 The z-score for the area of .7642 is 0.72 𝒛=𝟎.𝟕𝟐

a) What height represents the 95th percentile? We are given the mean 𝜇=64.1 and the standard deviation 𝜎=2.71 We need to find the x-value. We also need to find the z-score. The 95th percentile converts to an area of .9500 The area in the table closest to .9500 is .9505, which is a z-score of 1.65

Use the following formula to find the value, x: 𝑥=𝜇+𝑧𝜎 𝑥=64.1+(1.65)(2.71) 𝑥=64.1+4.4715 𝑥=68.5715 inches 𝒙=𝟔𝟖.𝟔 𝐢𝐧𝐜𝐡𝐞𝐬

b) What height represents the first quartile? We are given the mean 𝜇=64.1 and the standard deviation 𝜎=2.71 We need to find the x-value. We also need to find the z-score. The first quartile converts to an area of .2500 The area in the table closest to .2500 is .2514, which is a z-score of −0.67

𝑥=𝜇+𝑧𝜎 𝑥=64.1+(−0.67)(2.71) 𝑥=64.1−1.8157 𝑥=62.2843 inches Use the following formula to find the value, x: 𝑥=𝜇+𝑧𝜎 𝑥=64.1+(−0.67)(2.71) 𝑥=64.1−1.8157 𝑥=62.2843 inches 𝒙=𝟔𝟐.𝟑 𝐢𝐧𝐜𝐡𝐞𝐬

𝝁 𝒙 =𝟏𝟒𝟒.𝟑 𝐩𝐨𝐮𝐧𝐝𝐬 𝝈 𝒙 = 𝝈 𝒏 = 𝟓𝟏.𝟔 𝟑𝟓 =𝟖.𝟕𝟐 pounds Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution. Then sketch a graph of the sampling distribution. The consumption of processed fruits by people in the United States in a recent year was normally distributed, with a mean of 144.3 pounds and a standard deviation of 51.6 pounds. Random samples of size 35 are drawn from this population. 𝝁 𝒙 =𝟏𝟒𝟒.𝟑 𝐩𝐨𝐮𝐧𝐝𝐬 𝝈 𝒙 = 𝝈 𝒏 = 𝟓𝟏.𝟔 𝟑𝟓 =𝟖.𝟕𝟐 pounds

118.14 126.86 135.58 144.3 153.02 161.74 170.46

I’m not going to school today.

Find the probability for the sampling distribution. The mean annual salary for chauffeurs is $29,200. A random sample of size 45 is drawn from this population. What is the probability that the mean annual salary is less than $29,000? Assume that 𝜎=$1500 𝜇 𝑥 =$29,200 𝜎 𝑥 = 𝜎 𝑛 = 1500 45 =223.61 𝑧= 𝑥 −𝜇 𝜎 𝑛 = 29000−29200 223.61 = −200 223.61 =−0.89 The probability that matches up with 𝑧=−0.89 is 0.1867 𝑷( 𝒙 <$𝟐𝟗,𝟎𝟎𝟎)≈𝟎.𝟏𝟖𝟔𝟕

Find the probability for the sampling distribution. The mean annual salary for chauffeurs is $29,200. A random sample of size 45 is drawn from this population. What is the probability that the mean annual salary is more than $29,000? Assume that 𝜎=$1500 𝜇 𝑥 =$29,200 𝜎 𝑥 = 𝜎 𝑛 = 1500 45 =223.61 𝑧= 𝑥 −𝜇 𝜎 𝑛 = 29000−29200 223.61 = −200 223.61 =−0.89 The probability that matches up with 𝑧=−0.89 is 0.1867 𝑷 𝒙 >$𝟐𝟗,𝟎𝟎𝟎 ≈𝟏−𝟎.𝟏𝟖𝟔𝟕≈𝟎.𝟖𝟏𝟑𝟑

A binomial experiment is given A binomial experiment is given. Decide whether you can use the normal distribution to approximate the binomial distribution. If you can, find the mean and standard deviation. If you cannot, explain why. A survey indicates that 59% of men purchased perfume in the past year. You randomly select 15 men and ask them if they have purchased perfume in the past year. 𝒏𝒑≥𝟓 𝟏𝟓 .𝟓𝟗 =𝟖.𝟖𝟓≥𝟓 𝒏𝒒≥𝟓 𝟏𝟓 .𝟒𝟏 =𝟔.𝟏𝟓≥𝟓 We can use the normal distribution to approximate the binomial distribution 𝝁=𝒏𝒑=𝟖.𝟖𝟓 𝝈= 𝒏𝒑𝒒 = 𝟏𝟓 .𝟓𝟗 .𝟒𝟏 = 𝟑.𝟔𝟑 ≈𝟏.𝟗𝟎

A binomial experiment is given A binomial experiment is given. Decide whether you can use the normal distribution to approximate the binomial distribution. If you can, find the mean and standard deviation. If you cannot, explain why. In a recent year, the American Cancer Society said that the five-year survival rate for new cases of stage 1 kidney cancer is 95%. You randomly select 12 men who were new stage 1 kidney cancer cases this year and calculate their five-year survival rate. 𝒏𝒑≥𝟓 𝟏𝟐 .𝟗𝟓 =𝟏𝟏.𝟒≥𝟓 𝒏𝒒≥𝟓 𝟏𝟐 .𝟎𝟓 =𝟎.𝟔𝟎<𝟓 Because one of the conditions is not met, we cannot use the normal distribution to approximate the binomial distribution.

Decide whether you can use the normal distribution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probability. Five percent of workers in the United States use public transportation to get to work. You randomly select 250 workers and ask them if they use public transportation to get to work. Find the probability that exactly 16 workers will say yes. 𝒏𝒑≥𝟓 𝟐𝟓𝟎 .𝟎𝟓 =𝟏𝟐.𝟓≥𝟓 𝒏𝒒≥𝟓 𝟐𝟓𝟎 .𝟗𝟓 =𝟐𝟑𝟕.𝟓≥𝟓 Because both conditions are met, we can use the normal distribution to approximate the binomial distribution.

We need to find the mean of our binomial distribution. 𝑛𝑝= 250 .05 =12.5 We need to find the standard deviation of our binomial distribution. 𝜎= 𝑛𝑝𝑞 = 250 .05)(.95) = 11.875 =3.45 We now need to set up the correction for continuity. 15.5<𝑥<16.5 In order to get the probability, we need a z-score. We have the following components: 𝑥=15.5 and 16.5 𝜇=12.5 𝜎=3.45

𝑥=15.5 and 16.5 𝜇=12.5 𝜎=3.45 We will use the z-score formula to help find the probability. 𝑧= 𝑥−𝜇 𝜎 𝑧= 15.5−12.5 3.45 = 3 3.45 =0.87 𝑧= 16.5−12.5 3.45 = 4 3.45 =1.16

The probability to the left at 𝑧=0.87 is .8078 Subtract the two probabilities. .8770−.8078=.0692 The probability that exactly 16 workers will say yes is .0692

Decide whether you can use the normal distribution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. Five percent of workers in the United States use public transportation to get to work. You randomly select 250 workers and ask them if they use public transportation to get to work. Find the probability that at least 9 workers will say yes. We know the mean is 12.5 and the standard deviation is 3.45. The correction for continuity is 𝑥>8.5 𝑧= 𝑥−𝜇 𝜎 = 8.5−12.5 3.45 = −4 3.45 =−1.16 The probability to the left of the z-score at −1.16 is .1230 The probability that at least 9 workers will say yes is 𝟏−.𝟏𝟐𝟑𝟎=.𝟖𝟕𝟕𝟎

Cat Burglars