Question 1172802: A management consultant found thathe amount of
time per day spent by executives performing tasks
that could be done equally well by subordinates fol-
lowed a normal distribution with a mean of 2.4 hours
It was also found that 10% of executives spent over
3.5 hours per day on tasks of this type. For a random
sample of 400 executives, find the probability that
more than 80 spend more than 3 hours per day on
tasks of this type.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
1.28=(3.5-2.4)/sd
sd=1/1/1.282=0.859 hours
the 1.28 is the 90th percentile of the curve with mean 2.4 and sd 0.859
Find the z-value for 3 hours which is 0.6/0.859 or 0.698
This corresponds to 0.2425 probability for a given individual. This means that from 400 individuals, we would expect 97 to spend more than 3 hours a day, so the probability should be quite high that more than 80 do.
do a 1 proportion z-test with z=(phat-p)/sqrt (p*(1-p)/n) and phat =80/400=0.20
this is (0.20-0.2425)/sqrt(0.2425*0.7575)/400
=-0.0425/0.02143
=-1.983
probability z >= -1.963 is 0.9763
Can also do a normal approximation to this value. np=mean or 0.2425*400=97
variance is np(1-p) or 97(0.7575)=73.48
sd is sqrt (V)=8.57
z=(80.5-97/8.57, more than 80 with continuity correction factor.
z=-16.5/8.57
=-1.925
probability is 0.9729, so reasonable check.
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