Adam is a collegiate high jumper. The heights of his jumps follow a normal distr

Question

Adam is a collegiate high jumper. The heights of his jumps follow a normal distribution with an average of 6 feet and a standard deviation of 1.36 feet. Let Y be the height Adam can clear on his next jump. (Note: high jumpers try to jump over a bar that is set at a particular height for an attempt. Clearing the bar means that the high jumper went over the bar without knocking it down.) a) The bar is currently set at 7 feet. What is the probability that Adam does not clear the bar? b) What is the probability that Adam jumps at least 9 feet in his attempt? c) What is the 80th percentile of Adam’s jump heights? d) Adam has already done a few jumps in a competition and is very tired. He knows that on his next jump he can only jump at most 7 feet. If the bar is set at 5.87 feet, what is the probability that he clears it?

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    7      
u = mean =    6      
          
s = standard deviation =    1.36      
          
Thus,          
          
z = (x - u) / s =    0.735294118      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0.735294118   ) = P(x<7) =   0.768919805 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    9      
u = mean =    6      
          
s = standard deviation =    1.36      
          
Thus,          
          
z = (x - u) / s =    2.205882353      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.205882353   ) =    0.013696119 [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.8      
          
Then, using table or technology,          
          
z =    0.841621234      
          
As x = u + z * s,          
          
where          
          
u = mean =    6      
z = the critical z score =    0.841621234      
s = standard deviation =    1.36      
          
Then          
          
x = critical value =    7.144604878   [ANSWER]

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d)

Note that

P(x>5.87|x<7) = P(5.87<x<7)/P(x<7).

For P(5.87<x<7):

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    5.87      
x2 = upper bound =    7      
u = mean =    6      
          
s = standard deviation =    1.36      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.095588235      
z2 = upper z score = (x2 - u) / s =    0.735294118      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.461923805      
P(z < z2) =    0.768919805      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.306996

Hence,

P(5.87<x<7) = 0.306996

Thus,

P(x>5.87|x<7) = P(5.87<x<7)/P(x<7) = 0.306996/0.768919805 = 0.399256201 [ANSWER]

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