we saw two distributions for triathlon times:N(μ= 4313,σ=583) forMen, Ages 30 - 34andN(μ= 5261,σ= 807) for theWomen, Ages 25 - 29group. Times are listedin seconds. Use this information to compute each of the following:
(a) The cutoff time for the fastest 5% of athletes in the men’s group, i.e. those who took the shortest 5%of time to finish.
(b) The cutoff time for the slowest 10% of athletes in the women’s group.

Let X be the cut-off time for the fastest 5% of athletes in the men's group and [tex]z_{x}[/tex] be the corresponding z-score of X in the distribution N(μ= 4313,σ=583).

Then P(z<[tex]z_{x}[/tex])=0.05 and [tex]z_{x}[/tex]=-1.645

z score can be calculated using the formula

[tex]z_{x}[/tex]=1.645=[tex]\frac{X-M}{s}[/tex] where

X is the cut-off time for the fastest 5%
M is the mean time in seconds (4313)
s is the standard deviation (583)

putting the numbers in the formula we get

-1.645=[tex]\frac{X-4313}{583}[/tex] , solving for X in the equation we get

X=3353.965

(b) The cutoff time for the slowest 10% of athletes in the women’s group

Let Y be the cut-off time for the slowest 10% of athletes in the women's group and [tex]z_{y}[/tex] be the corresponding z-score of Y in the distribution N(μ= 5261,σ= 807).

Then P(z<[tex]z_{y}[/tex])=0.90 and [tex]z_{y}[/tex]=1.282

Using the above formula

putting the numbers in the formula we get

1.282=[tex]\frac{Y-5261}{807}[/tex] , solving for Y in the equation we get

Y=6295.574