Question

Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.

(a) Suppose n = 28 and p = 0.18. Can we approximate p̂ by a normal distribution? Why? (Use 2 decimal places.)

np = nq =

, p̂ be approximated by a normal random variable because .

What are the values of 𝜇_p̂ and 𝜎_p̂? (Use 3 decimal places.) 𝜇_p̂ = 𝜎_p̂ =

(b) Suppose n = 25 and p = 0.15. Can we safely approximate p̂ by a normal distribution? Why or why not?

, p̂ be approximated by a normal random variable because .

(c) Suppose n = 40 and p = 0.39. Can we approximate p̂ by a normal distribution? Why? (Use 2 decimal places.)

np = nq =

, p̂ be approximated by a normal random variable because .

What are the values of 𝜇_p̂ and 𝜎_p̂? (Use 3 decimal places.) 𝜇_p̂ = 𝜎_p̂ =

Answer 1

If $X\sim Bin(n, p)$and if $n$ is large and/or $p$ is close to $1/2,$ then $X$ is approximately $N(np, npq),$ where $q=1-p.$

In practice, the approximation is adequate provided that both $np\geq10$ and $nq\geq 10,$ since there is then enough symmetry in the underlying binomial

distribution.

Sometimes consider the condition $np>5$ and $nq>5.$

If the binomial probability histogram is not too skewed, $X$ has approximately a normal distribution with $\mu=np$ and $\sigma=\sqrt{npq}$

(a) $n=28, p=0.18$

$q=1-p=1-0.18=0.82$

Check the condition $np>5$ and $nq>5$

$np=28(0.18)=5.04>5$

$nq=28(0.82)=22.96>5$

We can safely approximate $\hat{p}$  by a normal distribution

$$\mu=np=28(0.18)=5.04$$

$$\sigma=\sqrt{npq}=\sqrt{28(0.18)(0.82)}=2.033$$

Check the condition $np\geq10$ and $nq\geq10$

$np=28(0.18)=5.04<10$

We cannot safely approximate $\hat{p}$  by a normal distribution.

(b) $n=25, p=0.15$

$q=1-p=1-0.15=0.85$

Check the condition $np>5$ and $nq>5$

$np=25(0.15)=3.75<5$

We cannot safely approximate $\hat{p}$  by a normal distribution.

Check the condition $np\geq10$ and $nq\geq10$

$np=25(0.15)=3.75<10$

We cannot safely approximate $\hat{p}$  by a normal distribution.

(c) $n=40, p=0.39$

$q=1-p=1-0.39=0.61$

Check the condition $np>5$ and $nq>5$

$np=40(0.39)=15.6>5$

$nq=40(0.61)=24.4>5$

We can safely approximate $\hat{p}$  by a normal distribution

$$\mu=np=40(0.39)=15.6$$

$$\sigma=\sqrt{npq}=\sqrt{40(0.39)(0.61)}=3.085$$

Check the condition $np\geq10$ and $nq\geq10$

$np=40(0.39)=15.6\geq10$

$nq=40(0.61)=24.4\geq10$

We can safely approximate $\hat{p}$ by a normal distribution

$$\mu=np=40(0.39)=15.6$$

$$\sigma=\sqrt{npq}=\sqrt{40(0.39)(0.61)}=3.085$$