(-6, -6) and (-2,-2)
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Let be the random variable for the number of marks a given student receives on the exam.
10% of students obtain more than 75 marks, so
where follows a standard normal distribution. The critical value for an upper-tail probability of 10% is
where denotes the CDF of
, and
denotes the inverse CDF. We have
Similarly, because 20% of students obtain less than 40 marks, we have
so that
Then are such that
and we find
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Let X be the number of lightning strikes in a year at the top of particular mountain.
X follows Poisson distribution with mean μ = 3.8
We have to find here the probability that in randomly selected year the number of lightning strikes is 0
The Poisson probability is given by,
P(X=k) =
Here we have X=0, mean =3.8
Hence probability that X=0 is given by
P(X=0) =
P(X=0) =
P(X=0) = 0.0224
The probability that in a randomly selected year, the number of lightning strikes is 0 is 0.0224