Question

By some estimates, thirty-­‐percent (30%) of all students in Groningen go on ski vacation each year. Out of a random sample of 300 students, what is the approximate probability that more than 100 of them went to ski this year? (Use the normal approximation)

And:

Oscar is getting married in Brazil where it rains only 1% of the time. Unfortunately, the weather expert has predicted rain for tomorrow. On a given day, there is a 9% chance that the weather expert predicts rain. 90% of the time that the weather expert has forecasted rain, it does in fact rain. What is the probability that it will rain on the day of Oscar´s wedding?

Answer 1

Answer 1

proportion p = 0.30

sample size n = 300

Using the normal approximation, we know that mean is given as

Mean = n*p = 300*0.3 = 90

Standard deviation = sqrt(n*p*(1-p)). Sqrt is square root here

Standard deviation = sqrt(300*0.3*(1-0.3)) = sqrt(63)

this gives us standard deviation = 7.94

We have to find the probability that mor than 100 of them went to ski this year

z calculation

z (x-mean)/(standard deviation )

putting x =100, mean = 90 and standard deviation = 7.94

So, z = (100-90)/(7.94)

= (10/7.94)

or z = 1.26

Using z distribution to find the value of P(more than 1.26)

we get p value = 0.1038 (In z distribution, find 1.2 in first column and 0.06 in first row, select the intersecting cell)

So, there is 0.1038 probability that more than 100 of them went to ski this year.