Question

In Assignment 2 you considered the following setting: Near-sightedness (myopia) afflicts roughly 10% of children at age 5.

Assume that the school has a total of 600 students.

Part A. In this school, what is the expected number of students to have myopia? What is the variance?

Part B. If we assume that the students in the school form an independent and random sample, then approximate the probability that at most 60 of the students in the 600 student school have myopia.

When calculating the approximate probability, please be sure to justify the approximation (i.e. check any conditions) and show all your work in the calculation.

Part C. If we assume that the students in the school form an independent and random sample, then what is the approximate probability that at least 55 and less than 65 of the students in the 600 student school have myopia? Again, show all necessary work in this calculation.

Answer 1

Given:  Binomial distribution with parameters n and p

p = proportion of student to have myopia = 10% = 0.10

n = total number of students = 600

Part A:

Expected number of students to have myopia = n * p = 600 * 0.10 = 60

Variance = n * p * (1 - p) = 600 * 0.10 *(1 - 0.10) = 600 * 0.10 * 0.90 = 54

Part B:

Here n * p = 60 and n * (1 - p) = 540 both are grreater than 5, so here the Normal approximation to the Binomial is applicable.

P(at most 60) that is

Binomial distribution is discrete and the Normal is conitnuous distribution, so here to find the probability need to do the continuity correction.

So here

First convert X into z score, the formula of z score is

P(X < 60.5) becomes P(Z < 0.07)

By using z table the probability for z = 0.07 is 0.5279

Therefore, the approximate probability that at most 60 of the students in the scholl have myopia is 0.5279

Part C :

P(at least 55 and less then 65) that is

after continuity correction the probability becomes P(54.5 < X < 64.5)

The z score for X = 54.5

z score for X = 64.5

The probability for z = -0.75 is 0.2266 and for z = 0.61 is 0.7291

So the required probability is 0.7291 - 0.2266 = 0.5025

Therefore, the approximate probability that at least 55 and less than 65 of the students in the school have myopia is 0.5025