The mean clotting time of blood is 7.45 seconds with a standard deviation of 3.6 seconds. What is the probability that an individual's clotting time will be less than 6 seconds or greater than 11 seconds? Assume a normal distribution. The probability is nothing.

The probability that an individual'c clothing time will be less than 6 seconds or greater than 11 is 0.509

Step-by-step explanation:

Given

Mean, μ = 7.45

Standard deviation, σ = 3.6

To solve this problem;

P(X<6 or X>11) = P(X<6) + P(X>11)

First, the z score of the individual clothing needs to be calculated

z score is calculated using (x - μ)/σ

where x = 6 seconds or 11 seconds

when x = 6

z = (x - μ)/σ

z = (6 - 7.45)/3,6

z = - 1.45/3,6

z = -0.403

when x = 11

z = (x - μ)/σ

z = (11 - 7.45)/3,6

z = 3.5/3,6

z = 0.972

So,

P(X<6 or X>11) = P(z<-0.403) + P(z>0.972) ---using z table

P(X<6 or X>11) = P(z<-0.403) +1 - P(z<=0.972)

P(X<6 or X>11) = 0.343 +1 - P(z<=0.972)

P(X<6 or X>11) = 0.343 + 1 - 0.834

P(X<6 or X>11) = 0.509.

Hence, the probability that an individual'c clothing time will be less than 6 seconds or greater than 11 is 0.509

stops on an even number.

step-by-step explanation:

4227473658779875563886538705958641728563470543621287463824132573847969p05749365273847906970680574936214