Question

Helen has an early class tomorrow morning. She knows that she needs to get to bed by 10:00 p.m. in order to be sufficiently rested to concentrate ancpate in class. However, before she goes to bed, she must start and then complete a homework assignment which is due tomorrow morning. According to her experience, the time it takes her to complete a homework assignment for this class is Normally distributed with mean u = 3.5 hours and standard deviation S = 1.2 hours. Helen looks at her watch and sees that it is now 6:00 PM. What is the probability that she will be able to get to bed in time to be sufficiently rested for the next day's class?

Answer 1

The probability that she will be able to get to bed in time is 66%

Explanation:

Hi!

The solution of this problem is given by the cumulative disrtibution of the normal distribution, we must integrate that pdf from -∞ to 4 hours, which is the time until its 10:00 pm.

Anologously we can use the Z score needed and look for tables and know the probability.

The Z score is given by:

Z = (x- m)/σ

Z = (4 -3.5)/1.2 =0.416

In the tables we find that for:

Z = 0.41 ----> P=0.65910

Z = 0.42 ----> P=0.66276

Therefore, if we consider only two significant figures we can say that:

The probability that she will be able to get to bed in time is 66%