Women have pulse rates that are normally distributed with a

Women have pulse rates that are normally distributed with a mean of 73.6 beats per minute and a standard deviation of 11.5 beats per minute.

Find the percentiles p1 and p99

Solution

P1:

P(X<c)=0.01

--> P((X-mean)/s <(c-73.6)/11.5) =0.01

--> P(Z<(c-73.6)/11.5) =0.01

--> (c-73.6)/11.5= -2.33 (from standard normal table)

So c= 73.6-2.33*11.5=46.805

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P99:

P(X<c)=0.99

--> P((X-mean)/s <(c-73.6)/11.5) =0.99

--> P(Z<(c-73.6)/11.5) =0.99

--> (c-73.6)/11.5= 2.33 (from standard normal table)

So c= 73.6+2.33*11.5=100.395