For a recent 10k run the finishers are normally distributed

For a recent 10k run the finishers are normally distributed with mean 62 minutes and standard deviation 12 minutes. a. Determine the percentage of finishers with times between 45 and 75 minutes. b. Find the percentage of finishers with times less than 80 minutes. c. Obtain the 35th percentile for the finishing times. d. Find the seventh decile for the finishing times. a. Approximately % of finishers had times between 45 and 75 minutes. (Round to two decimal places as needed.) b. Approximately % of finishers had times less than 80 minutes. (Round to two decimal places as needed.) c. The 35th percentile is minutes. (Round to two decimal places as needed.) d. The seventh decile is minutes. (Round to two decimal places as needed.)

Solution

A.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    45      
x2 = upper bound =    75      
u = mean =    62      
n = sample size =    1      
s = standard deviation =    12      
          
Thus, the two z scores are          
          
z1 = lower z score =    -1.416666667      
z2 = upper z score =    1.083333333      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.078290204      
P(z < z2) =    0.860669753      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.782379549 = 78.24%

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b.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    80      
u = mean =    62      
n = sample size =    1      
s = standard deviation =    12      
          
Thus,          
          
z =    1.5      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   1.5   ) =    0.933192799 = 93.32%

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C.

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.35      
          
Then, using table or technology,          
          
z =    -0.385320466      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    62      
z = the critical z score =    -0.385320466      
s = standard deviation =    12      
n = sample size =    1      
Then          
          
x = critical value =    57.3761544   [ANSWER]

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First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.7      
          
Then, using table or technology,          
          
z =    0.524400513      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    62      
z = the critical z score =    0.524400513      
s = standard deviation =    12      
n = sample size =    1      
Then          
          
x = critical value =    68.29280615   [ANSWER]