The times of the finishers in the New York City 10 km run ar

The times of the finishers in the New York City 10 km run are normally distributed with mean 61 minutes and standart deviation 8.4 minutes.

a)Find the probability taht a randomly selected finisher does the race in under 55 minutes

b)Find the probability that a randomly selected finisher completes the race in between 62 and 85 minutes

c)Find the 30th percentile

d)If 9 different runners are randomly selected, find the probability that the nine runners have a mean finishing time under 55 minutes.

Solution

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    55      
u = mean =    61      
          
s = standard deviation =    8.4      
          
Thus,          
          
z = (x - u) / s =    -0.714285714      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.714285714   ) =    0.237525262 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    62      
x2 = upper bound =    85      
u = mean =    61      
          
s = standard deviation =    8.4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.119047619      
z2 = upper z score = (x2 - u) / s =    2.857142857      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.547381185      
P(z < z2) =    0.997862633      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.450481448   [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.3      
          
Then, using table or technology,          
          
z =    -0.524400513      
          
As x = u + z * s,          
          
where          
          
u = mean =    61      
z = the critical z score =    -0.524400513      
s = standard deviation =    8.4      
          
Then          
          
x = critical value =    56.59503569   [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    55      
u = mean =    61      
n = sample size =    9      
s = standard deviation =    8.4      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.142857143      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.142857143   ) =    0.016062286 [ANSWER]