Question

According to a recent study, 80% of American children (age 2 - 9) believe in Santa Claus. Suppose we obtain a random sample of 110 children in this age group and ask them whether or not they believe in Santa Claus. Answer the following:

1) Can the sampling distribution of ˆ p p ^ be approximated by the normal distribution? no yes

2) What is the mean of the distribution of ˆ p p ^ ?

3) What is the standard deviation of the distribution of ˆ p p ^ ?

4) What is the probability that fewer than 96 of the children from the sample will respond that they believe in Santa Claus?

(Round your answers to four decimal places when appropriate)

Answer 1

Solution:

We are given

Population proportion = p = 80% = 0.80

Sample size = n = 110

1) Can the sampling distribution of ˆ p p ^ be approximated by the normal distribution?

Answer: Yes, because we know that the sampling distribution of any sample statistic follows an approximately normal distribution.

2) What is the mean of the distribution of p̂?

Mean = n*p = 110*0.80 = 88

Mean = 88

3) What is the standard deviation of the distribution of p̂?

Standard deviation = sqrt(p̂*(1 - p̂)/n)

Standard deviation = sqrt(0.80*(1 – 0.80)/110)

Standard deviation = sqrt(0.80*0.20/110)

Standard deviation = sqrt(0.001455)

Standard deviation = 0.038139

4) What is the probability that fewer than 96 of the children from the sample will respond that they believe in Santa Claus?

Here, we have to use normal approximation to binomial distribution.

We have to find P(X<96)

We have n = 110, p = 0.80, q = 1 – p = 1 – 0.80 = 0.20

Mean = np = 110*0.80 = 88

SD = sqrt(npq) = sqrt(110*0.80*0.20) = 4.195235

P(X<96) ≈ P(X<95.5) (by using continuity correction)

Z = (X – mean) / SD

Z = (95.5 - 88)/ 4.195235

Z = 1.78774

P(Z<1.78774) = P(X<96) = 0.963091

(by using z-table)

Required probability = 0.9631