Answer:
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