Question

according to a report, 70.5% of murders are committed with a firearm
(a) if 300 murders are randomly selected, how many would we expect to be committed with a firearm?
(b) would it be unusual to observe 221 murders by firearm in a random sample of 300 murders? why?
(a) we would expect 211.5 to be committed with a firearm.
(b) choose the correct answer below
o a. no, because 221 is between μ - 2σ and μ + 2σ
o b. yes, because 221 is between μ - 2σ and μ + 2σ
o c. no, because 221 is greater than μ + 2σ
o d. no, because 221 is less than μ - 2σ
o e. yes, because 221 is greater than μ + 2σ

Explanation:

Step1: Calculate the mean ($\mu$)

The mean of a binomial distribution is given by $\mu = np$, where $n = 300$ (number of trials) and $p=0.705$ (probability of success).
$$\mu=300\times0.705 = 211.5$$

Step2: Calculate the standard deviation ($\sigma$)

The standard deviation of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$.
Substitute $n = 300$ and $p = 0.705$:

$$\begin{align*} \sigma&=\sqrt{300\times0.705\times(1 - 0.705)}\\ &=\sqrt{300\times0.705\times0.295}\\ &=\sqrt{62.3925}\\ &\approx7.90 \end{align*}$$

Step3: Find $\mu\pm2\sigma$

Calculate $\mu - 2\sigma$ and $\mu+2\sigma$:
$$\mu - 2\sigma=211.5-2\times7.90=211.5 - 15.8=195.7$$
$$\mu+2\sigma=211.5 + 2\times7.90=211.5+15.8 = 227.3$$

Since $195.7<221<227.3$ (i.e., $221$ is between $\mu - 2\sigma$ and $\mu+2\sigma$), the value is not unusual.

Answer:

A. No, because 221 is between $\mu - 2\sigma$ and $\mu+2\sigma$