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# Examples for Ch 7

### Elementary Statistics

If the scores for a test have a mean of 100 and a standard deviation of 15, find the percentage of scores that will fall below 112. X= 112, μ= 100, σ= 15; z=(X-μ)/σ =(112-100)/15 = 12/15 = 0.8; area = .7881, 78.81% of the scores will fall below 112.
Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds. Find the propility of its generating between 27 and 31 pounds per month. X₁= 27, X₂=31, μ= 28, σ= 2; z₁=(X₁-μ)/σ =(27-28)/2 = 3/2 = -0.5; z₂=(X₂-μ)/σ =(31-28)/2 = 1.5; area = .6247; There a 62.47% propability that a household will generate between 27 and 31 pounds of newspapers.
The average time it takes to respond to an emergency call is 25 minutes with a standard deviation of 4.5 minutes. If 80 calls are randomly selected, approximately how many will be respond to in less than 15 minutes? X= 15, μ= 25, σ= 4.5; z=(X-μ)/σ =(15-25)/4.5 = -2.22; area = .0132; n*area = 80(.0132)=1.056; Therefore, approximately one call will be responded to in under 15 minutes.
In order to qualify for a police academy, candidates must scorein the top 10% on a general abilities test. The test has a mean of 200 and a standard deviation of 20. Find the lowest possible score to qualify. 10%= .1= .1000= area; 1-area= 1-.1000= .9000; z=1.28; X= ?, μ= 200, σ= 20; z=(X-μ)/σ; 1.28= (X-200)/20; X = 226; A score of 226 is the cutoff.
The average age of a vehicle registered in the United States is 8 years, or 96 months. Standard deviation is 16 months. If a sample of 36 vehicles is selected, find the propability that the mean of their age is between 90 and 100 months. ¯X₁=90, ¯X₂=100, μ= 96, σ= 16, n=36; z=(¯X- μ)/(σ/√n); z₁=(90-96)/(16/√36)=-2.25; z₂=(100-96)/(16/√36)=1.50; area = 0.921; The propability of obtaining a sample mean between 90 and 100 months is 92.1%.
If a baseball player's batting average is 0.320 (32%), find the propability that the player will get at most 26 hits in 100 times at bat. X= 26, p=0.32, q=1-p=1-.32=.68, n=100; μ=np=(100)(.32)=32; σ=√npq= √(100)(.32)(.68)=4.66; P(X≤26), P(X<26+.5)=P(X<26.5); z=(X-μ)/σ = (26.5-32)/4.66=-1.18; area = .1190 or 11.9%.
q= Make sure p is in decimals: 1-p = q
Find the area under the normal distribution curve between z= 0 and z=2.34. Area = .4904
Find the area right of z=1.11. Area for z=1.11 is .8665; 1-.8665= .1335 or 11.35%.
Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds. Find the propility of its generating more than 30.2 pounds per month. X= 30.2, μ= 28, σ= 2; z=(X-μ)/σ =(30.2-28)/2 = 2.2/2 = 1.1; area = .1357; There a 13.57% propability that a household will generate more than 30.2 pounds of newspapers.
Created by: dengler