Question

Question 4 10 points A stunt person injures himself an average of three times a year. The probability that he will be injured More than three times this year is: O 0.6472 O 0.4232 0.3528 O 0.5768

Answer 1

4) Let the probability of getting injured x times a year be represented as P(x).

Given, a person injures himself an average of three times per year, so, $\mu$ =3.

Since, the average number of injuries happening per year is given, and we are asked to calculate a probability of getting injured more than 3 times in an year, so, the Poisson Distribution is used,so,

P(x)=$\frac{3^xe^{-3}}{x!}$

Probability of getting injured more than 3 times a year = P(x>3) = 1-P(0)-P(1)-P(2)-P(3)= $1-\frac{3^{0}e^{-3}}{0!}-\frac{3^{1}e^{-3}}{1!}-\frac{3^{2}e^{-3}}{2!} -\frac{3^{3}e^{-3}}{3!}$ = $1-\frac{e^{-3}}{1}-\frac{3e^{-3}}{1}-\frac{9e^{-3}}{2} -\frac{27e^{-3}}{6}$ ==

9e-3 9e -3 1-е-3 Зе-3 2 2

$1-13e^{-3}$ =0.3528.

Hence, 3rd option is the correct answer.

5)Mean=$\mu$=90

Standard deviation =$\sigma$=10

x=100

Z=$\frac{x-\mu}{\sigma}=\frac{100-90}{10}=\frac{10}{10}=1$

Probability that a car picked at random is travelling at more than 100 km per hour = P(x>100) = P(Z>1) = 0.1587. (By looking at the z-score table, we know for Z>1, P(Z>1)=0.1587.)

Hence, the first option is the correct answer.