Normal distribution
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A university professor keeps records of his travel time while he is driving between his home and the university. Over a long period of time he has found that his morning travel times are approx Normally distributed with a mean of 31 minutes and a standard deviation of 3.0 minutes;his return journey in the evening is similarly distributed but with a mean of 35.5 minutes and a standard deviation of 3.5 minutes.
a) Find the probability that on a typical day he spends more than one hour traveling to and from work.
b) Find the probability that on a given day his morning journey is longer than his evening journey?
c) On what proportion of days is the evening journey more than 5 minutes longer than the morning journey?
d) A new route was recently opened. He has discovered that if he leaves home at 8:30am and uses this new route, on 88% of days he arrives at the university by 9:00am. If the times of travel are still approx Normally distributed with a standard deviation of 3.0 minutes, find the mean time for his morning trip using the new route.
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Solution Summary
The expert analyzes normal distribution functions. The probability that on a typical day he spends more than one hour traveling to and from work is computed.The solution answers the question(s) below.
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a) Find the probability that on a typical day he spends more than one hour traveling to and from work.
we calculate the distribution of the round journey.
Since this is normally distributed,
Mean(Round)=Mean(Morning) + Mean(evening)=31+35.5=66.5 minutes
Var(Round)=var(Morning) + var(evening)=3.0^2+3.5^2=21.25
Then Sd(Round)=SQRT(Var)= SQRT(21.25)=4.61 minutes
Now the z value is
Z=(66.5-60)/4.61= 1.410
From the z-table, we find the z value of 1.410 is 0.9207
Therefore, the probability of T > 60 ...
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