Question

Can you walk me through the second part of this question? I got the first part right, just dont understand the second.

A professor knows that her statistics students' final exam scores are approximately normally distributed with a mean of 78 and a standard deviation of 7.5.

In her class, an "A" is any exam score of 90 or higher.

Give your answers rounded to 4 places after the decimal.

a) What is the probability that a randomly selected student will get an A on the exam?

0.0548

This quarter there are 31 students in her class.

b) What is the probability that at least 4 students will score an "A" on the final exam?

Hint: What kind of distribution are we talking about now?

Answer 1

A) P(X > 90)

= P((X - )/> (90 - )/)

= P(Z > (90 - 78)/7.5)

= P(Z > 1.6)

= 1 - P(Z < 1.6)

= 1 - 0.9452

= 0.0548

B) n = 31

P = 0.0548

It is a biomass distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X > 4)

= 1 - P(X < 4)

= 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))

= 1 - (31C0 * (0.0548)^0 * (1 - 0.0548)^31 + 31C1 * (0.0548)^1 * (1 - 0.0548)^30 + 31C2 * (0.0548)^2 * (1 - 0.0548)^29 + 31C3 * (0.0548)^3 * (1 - 0.0548)^28)

= 1 - 0.9126

= 0.0874