Question

According to government data, 22% of American children under the age of 6 live in households with incomes less than the official poverty level. A study of learning in early childhood chooses a SRS of 300 children. Estimate the probability that between 60 and 63 children of the sample are from poverty households?

Answer 1

Answer:

The estimated probability that between 60 and 63 children of the sample are from poverty households is 0.1367.

Step-by-step explanation:

Let X = number of American children under the age of 6 live in households with incomes less than the official poverty level.

The probability of the random variable X occurring is, p = 0.22.

A random sample of children of size n = 300 is selected.

The random variable X follows a Binomial distribution with parameters n = 300 and p = 0.22.

A normal approximation to Binomial can be used to approximate the distribution of X if the following conditions are fulfilled:

• np ≥ 10

• n(1 - p) ≥ 10

Check:

$np=300\times 0.22=66>10\\n(1-p)=300\times (1-0.22)=234>10$

So, the random variable X approximately follows N (np, np(1 - p)).

Compute the probability of X between 60 and 63 as follows:

$P(60<X<63)=P(\frac{60-(300\times 0.22)}{\sqrt{300\times 0.22\times (1-0.22)}}<\frac{X-np}{\sqrt{np(1-p)}}<\frac{63-(300\times 0.22)}{\sqrt{300\times 0.22\times (1-0.22)}})$

                          $=P(-0.84<Z<-0.42)\\=P(Z<-0.42)-P(Z<-0.84)\\=P(Z<0.84)-P(Z<0.42)\\=0.7995-0.6628\\=0.1367$

*Use a z-table for the probability values.

Thus, the estimated probability that between 60 and 63 children of the sample are from poverty households is 0.1367.