Answer:
a) The probability that the sample average amount of time taken on each day is at most 11 min.
P(X>11) = 0.2546
Step-by-step explanation:
Step1:-
Given data in a normal distribution with mean value 9 min and standard deviation 3 min.
Let 'X' be the amount of time taken on each day
Given x = 11 , μ=9min and σ=3min
by using normal distribution
[tex]Z= \frac{x-mean}{S.D} = \frac{11-9}{3}[/tex]
Z = 0.66>0
Step2:-
The probability that the sample average amount of time taken on each day is at most 11 min.
P(X>11) = P(z>z₁) = 0.5-A(z₁) ( check diagram)
= 0.5 - A(0.66)
= 0.5 - 0.2454 ( check normal table)
= 0.2546
Conclusion:-
The probability that the sample average amount of time taken on each day is at most 11 min.
P(X>11) = 0.2546
