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?

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

Step-by-step 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%

step-by-step explanation:

f(gx)=f(1/x+2)

=(1/x+2)+2

=(1+2x)/2+2

=(1+2x+4)/2

= (2x+5)/2

hello,

as   long as they add up to 180 its correct.