Question

The amount of time each week that Dunder Mifflin employees spend in pointless meetings follows a normal distribution with a mean of 90 minutes and standard deviation of 10 minutes.

A. What is the probability that employees spend between 81.5 and 103.5 minutes in meetings in a week?

B. The probability that employees spend between 80 and ???? minutes in meetings in a week is equal to 0.8351.

Answer 1

Solution

Let X = amount of time (minutes) each week that Dunder Mifflin employees spend in pointless meetings.

We are given: X ~ N(90, 10²)

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then, pdf of X, f(x) = {1/σ√(2π)}e^-[(1/2){(x - µ)/σ}²] ………………………………………………………(1)

P(X ≤ t) is obtained by integrating f(x) given under (1) from - ∞ to t …............………….………(2)

The integral is quite difficult to evaluate. However, it is not necessary to go through the integration ordeal. The cumulative probability values for the Normal Variable, X ~ N(µ, σ²), can be found directly using Excel Function NORMDIST(x,Mean,Standard_dev,Cumulative)……………......………..(3)

Percentage points of N(µ, σ²) can be found using Excel Function: NORMINV(probability,mean,standard_dev ) which gives values of t for which P(X ≤ t) = given probability………………………………………………………………………………………..……(3a)

P(t1 ≤ X ≤ t2) = P(X ≤ t2) - P(X ≤ t1) ………………………………………………………………. (4)

Now to work out the solution,

Part (a)

Probability that employees spend between 81.5 and 103.5 minutes in meetings in a week

= P(81.5 ≤ X ≤ 103.5)

= P(X ≤ 103.5) - P(X ≤ 81.5) [vide (4)]

= 0.911492 – 0.197663 [vide (3)]

= 0.717663 ANSWER

Part (b)

Let t be the required value. Then, we are given: P(80 ≤ X ≤ t) = 0.8351

i.e., P(X ≤ t) - P(X ≤ 80) = 0.8351 [vide (4)]

=> P(X ≤ t) – 0.158655 = 0.8351 [vide (3)]

Or, P(X ≤ t) = 0.993755

=> t = 114.9799 [vide (3a)] ANSWER

DONE