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 (X11) = P (X11)

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 (X11) = P (z0.972) - --using z table

P (X11) = P (z<-0.403) + 1 - P (z<=0.972)

P (X11) = 0.343 + 1 - P (z<=0.972)

P (X11) = 0.343 + 1 - 0.834

P (X11) = 0.509.

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