1. 2. 3. 4. 5. The mean finish time for a yearly amateur auto race was 185.45 mi
ID: 3203786 • Letter: 1
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The mean finish time for a yearly amateur auto race was 185.45 minutes with a standard deviation of 0.309 minute. The winning car, driven by Hank, finished in 184.49 minutes. The previous years race had a mean finishing time of 112.1 with a standard deviation of 0.114 minute. The winning car that year, driven by Nina, finished in 82 minutes. Find their respective z-scores. Who had the more convincing victory? Hank had a finish time with a z-score of Nina had a finish time with a z-score of (Round to two decimal places as needed.) Which driver had a more convincing victory? O A. Hank had a more convincing victory because of a higher z-score. O B. Nina a more convincing victory because of a higher zscore. O C. Nina a more convincing victory because of a lower z-score. O D. Hank had a more convincing victory because of a lower z-score.Explanation / Answer
Problem 1) Hank mean = 185.45 , ,std. deviation = 0.309 , X = 184.49
By central limit theorem,
z = ( x - mean ) / Std. deviation
= ( 184.49 - 185 . 45 ) / 0.309
= -3.10
Hank z score = -3.10
Nina mean = 112.1 , std. deviation = 0.114 , X = 111.82
By central limit theorem,
z = ( x - mean ) / Std. deviation
= ( 111.82 - 112.1 ) / 0.114
= -2.45
Nina z score = -2.45
Nina has a more convincing victory because of higher z score.
Problem 3)
Mean = 300 , std. deviation = 20
a) 95% of organs will be within 2 standard deviations of the mean
= 300 - 2 (20) and 300+ 2(20)
= (260, 340)
b) 300 - 1 ( 20 ) and 300 + 1( 20 )
= 280 and 320
68% of observations lie within 1 standard deviations of the mean
c) 100 - 68% = 32 %
300 - 2 ( 20 ) and 300 + 1(20)
= 280 and 320
16 % of observations lie above 320 and 2.5% of observations lie below 280
d) 100 - 16 - 2.5 = 81.5%
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