Question

According to the Insurance Institute of America, a family of four spends between $400 and $3,800 per year on all types of insurance. Suppose the money spent is uniformly distributed between these amounts.
a. What is the mean amount spent on insurance?
b. What is the standard deviation of the amount spent?
c. If we select a family at random, what is the probability they spend less than $2,000 per year on insurance per year?
d. What is the probability a family spends more than $3,000 per year?

Answer 1

a. Because of the uniform distribution, the mean = the middle of both numbers, or 1/2(3800 + 400) = $2100.
b. To find the standard deviation, we take the variance ((b-a)/√12) and plug in the values:
(3800 - 400)/√12
3400/√12
~981.5
c. Again, because of uniform distribution, we just have to find the percentage that all numbers less than 2000 have in the range of 400 to 3800. To make things simple, I'm going to subtract 400 from both 3800 and 2000 so the data cuts off at 0. The equation is:
1600 = 3400 * x
16/34 = x (simplified from 1600/3400)
8/17 = x
8/17 is about 0.47, or 47%, so that's your amount.
d. same thing with c, but slightly different:
3800 - 400 = 3400, 3000 - 400 = 2600
3400 - 2600 = 800
800 = 3400 * x
8/34 = x
4/17 = x
4/17 is about 0.235.

Answer 2

Answer: a. 2100

b. 981.5

c. 0.471

d. 0.235

Step-by-step explanation:

Given: According to the Insurance Institute of America, a family of four spends between $400 and $3,800 per year on all types of insurance.

If the money spent is uniformly distributed between these amounts.

Let A=$400 and B= $3,800

a. The mean amount spent on insurance = $\frac{A+B}{2}=\frac{3800+400}{2}=\frac{4200}{2}=\ 2100$

b. The standard deviation of the amount spent=

$\sqrt{\frac{(B-A)^2}{12}}\\=\sqrt{\frac{(3800-400)^2}{12}}\\=\sqrt{\frac{(3400)^2}{12}}$

$=\frac{3400}{\sqrt{12}}=981.495\approx981.5$

c. To calculate, the probability they spend less than $2,000 per year on insurance per year ,it means the amount is between 400 and 2000 i.e

Amount=$2000-$400=$1600

Then P(400<insurance amount<2000)$=\frac{amount}{B-A}=\frac{1600}{3400}=0.4705\approx0.471$

d. Ia a family spends more than $3000 then the amount is between $3000 and $3800, i.e. Amount= $3800-$3000=$800

Now,

P(3000<insurance amount<3800)=$\frac{800}{3400}=\frac{4}{17}=0.235$