Lightning Deaths Based on the results given in Cumulative Review Exercise 6, assume that for a randomly selected lightning death, there is a 0.8 probability that the victim is a male.

a. Find the probability that three random people killed by lightning strikes are all males.

b. Find the probability that three random people killed by lightning strikes are all females.

c. Find the probability that among three people killed by lightning strikes, at least one is a male.

d. If five people killed by lightning strikes are randomly selected, find the probability that exactly three of them are males.

e. A study involves random selection of different groups of 50 people killed by lightning strikes. For those groups, find the mean and standard deviation for the numbers of male victims.

f. For the same groups described in part (e), would 46 be a significantly high number of males in a group? Explain

The probability of a male death due to lightning strikes is equal to 0.8.

a.

The probability of all 3 people killed being male is computed below:

$$\begin{gathered} P\left(\text{all}\text{three}\text{males}\right)=0.8\times 0.8\times 0.8 \\ =0.512 \end{gathered}$$

Thus, the probability of all 3 people killed being male is equal to 0.512.

b.

The probability of all 3 people killed being female is computed below:

$$\begin{gathered} P\left(\text{all}\text{three}\text{males}\right)=\left(1-0.8\right)\times \left(1-0.8\right)\times \left(1-0.8\right) \\ =0.008 \end{gathered}$$

Thus, the probability of all 3 people killed being female is equal to 0.008.

c.

The probability of at least one of the 3 people killed being male is computed below:

$$\begin{gathered} P\left(\text{at}\text{least}\text{one}\text{male}\right)=1-P\left(\text{all}\text{three}\text{females}\right) \\ =1-\left[\left(1-0.8\right)\times \left(1-0.8\right)\times \left(1-0.8\right)\right] \\ =1-0.008 \\ =0.992 \end{gathered}$$

Thus, the probability of at least one of the 3 people killed being male is equal to 0.992.

d.

Let p denote the probability of male death.

$p=0.8$

Let q denote the probability of female death.

$$\begin{gathered} q=1-p \\ =1-0.8 \\ =0.2 \end{gathered}$$

Let n denote the number of trials.

$n=5$

Let x denote the number of successes (number of male deaths).

$x=3$

Using the binomial probability formula, the probability of 3 male deaths is equal to:

$$\begin{gathered} P\left(X=x\right)={}^n{C}_x{\left(p\right)}^x{\left(q\right)}^{n-x} \\ P\left(X=3\right)={}^5{C}_3{\left(0.8\right)}^3{\left(0.20\right)}^{5-3} \\ =\frac{5!}{3!\left(5-3\right)!}{\left(0.8\right)}^3{\left(0.20\right)}^2 \\ =0.205 \end{gathered}$$

Thus, the probability of getting exactly male deaths out of 5 randomly selected deaths is equal to 0.205.Step 3: Binomial probability

e.

Let n be equal to 50.

The mean number of male deaths is equal to:

$$\begin{gathered} \mu =np \\ =\left(50\right)\left(0.8\right) \\ =40 \end{gathered}$$

Thus, the mean number of male deaths is equal to 40.

The standard deviation of the number of male deaths is equal to:

$$\begin{gathered} \sigma =\sqrt{npq} \\ =\sqrt{\left(50\right)\left(0.8\right)\left(0.2\right)} \\ =2.8 \end{gathered}$$

Thus, the standard deviation is equal to 2.8.

f.

Values that are significantly low are less than or equal to $\mu -2\sigma$.

$$\begin{gathered} \mu -2\sigma =40-2\left(2.8\right) \\ =34.4 \\ \approx 34 \end{gathered}$$

Values that are significantly high are greater than or equal to $\mu +2\sigma$.

$$\begin{gathered} \mu +2\sigma =40+2\left(2.8\right) \\ =45.6 \\ \approx 46 \end{gathered}$$

Values that lie between 34 and 46 are neither significantly low nor significantly high.

Here, the value of 46 is equal to the limit separating significantly high values.

Thus, the value of 46 male deaths is significantly high.