SOLUTION: A research concludes that the number of hours of exercise per week for adults is normally distributed with a mean of 3.5 hours and a standard deviation of 3 hours. Show all work. J

Question 1058647: A research concludes that the number of hours of exercise per week for adults is normally distributed with a mean of 3.5 hours and a standard deviation of 3 hours. Show all work. Just the answer, without supporting work, will receive no credit.
(a) Find the 80th percentile for the distribution of exercise time per week. (round the answer to 2 decimal places)
(b) What is the probability that a randomly selected adult has more than 7 hours of exercise per week? (round the answer to 4 decimal places)
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The 80th percentile is p=0.8000 on the std normal curve
z=0.842
z=(x-mean)/sd
0.842=(x-3.5)/3
2.526=x-3.5
x=6.026 or 6.03 hours. That is 2.53 above the mean or 0.84 sd or 80th percentile.
more than 7 hours
z=(7-3.5)/3=3.5/3=1.167
for z>1.167, the probability is 0.1216 or 0.12.
It is more than 1 sd, above which is about 15% of the probability, so this makes sense.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A research concludes that the number of hours of exercise per week for adults is normally distributed with a mean of 3.5 hours and a standard deviation of 3 hours. Show all work. Just the answer, without supporting work, will receive no credit.
(a) Find the 80th percentile for the distribution of exercise time per week. (round the answer to 2 decimal places)
sample mean:: 3.5
ME = 1.282*3 = 3.845
80% CI:: 3.5-3.845 < u < 3.5+3.845
80% CI: -0.345 < u < 7.345
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(b) What is the probability that a randomly selected adult has more than 7 hours of exercise per week? (round the answer to 4 decimal places)
z(7) = (7-3.5)/3 = 3.5/3 = 1.167
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P(x > 7) = P(z > 1.167) = normalcdf(1.167,100) = 0.1217
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Cheers,
Stan H.
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