a) He missed all the hints, that means he reached the class after 10:07
That means total time taken by him in reaching class is more than or equal to 27 minutes as he started at 9:40 and reached at or after 10:07
So, we need to find "P(X \\ge 27)"
First we convert it into Z score, then using normal distribution table, we will find the required probability.
So,
"Z= (x-\\mu)\/\\sigma"
Where the mean is 25 minutes and the standard deviation is 3 minutes.
Putting up the value, we will get
"Z=(27-25)\/3"
"=2\/3=0.667"
From the normal probability distribution table, it can be seen that "P(Z \\le 0.667)=0.748"
Hence the "P(Z \\ge 0.667)=1-0.748"
"=0.252"
b) Given that "P(Z \\le z)=0.97"
From the normal distribution table we can find the value of z to be 1.881
So let's suppose the time taken by him to reach office is x minutes.
Then,
"(x-25)\/3=1.881"
"x=(3*1.881)+25"
"x=30.643"
So, he must leave his home by 9:00 to reach the lecture by 9:30 with 97% probability.
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