Question

Approximately 54% of mathematics students do their homework on time. In a class of 250 students, what is the mean, variance, and standard deviation if we assume normality and use the normal distribution as an approximation of the binomial distribution? Answer choices rounded to the nearest whole number.

a.) Mean = 135
Variance = 62
Standard Deviation = 8

b.) Mean = 53
Variance = 62
Standard Deviation = 8

c.) Mean = 54
Variance = 8
Standard Deviation = 62

d.) Mean = 135
Variance = 8
Standard Deviation = 62

Answer 1

The problem above is related to Binomail Distribution

Given 54% of the maths students do home work on time,

54% in decimals = 54/100 = 0.54

So here p = 0.54 , p is the probability that a student does home work on time

p + q = 1, q = 1 - p = 1 -0.54 = 0.46

q = 0.46 , where q is the probability that a student does not do home work on time.

n = total sample size = 250,

Mean of a Binomail Distribution = np, n =250, p = 0.54

Mean = np = 0.54 ✕ 250 = 135

variance of a Binomail Distribution = npq = 250 ✕ 0.54 ✕ 0.46 = 62.1,

variance of a Binomail Distribution = 62 (rounded to nearest whole number)

OPTION A IS THE ANSWER