asked 179k views
0 votes
 Probability Distributions for Discrete Random Variables

A life insurance policy pays 1000 upon the death of a policyholder provided that the policyholder su at least one year but less than five years after purchasing the policy. Let X denote the number of yea policyholder survives after purchasing the policy with the following probabilities:

X 1 2 3 4 5
P(x). 0.05 0.12 0.21 0.33 0.48

Calculate the mean, the variance and the standard deviation of the payment made under this polic

 Probability Distributions for Discrete Random Variables A life insurance policy-example-1
asked
User Macil
by
8.5k points

1 Answer

3 votes

1. mean = average

2. variance means :

if everyone got a high or low score on a test then the variance is low or 0

variance means are there a lot of different test scores ?

if there are then the variance is high

variance = 1. each different number minus the average squared 2. add them all up 3. divide the total by the total number of numbers

3. square root of variance = standard deviation

standard deviation is like the distance between a person's house and their friend's house to see how far they typically have to travel.

answers :

mean= 4.64 , variance= -2.4596 , standard deviation = 1.57

steps

rewrite values

X, P(X)

1 , 0.05

2 , 0.12

3 , 0.21

4 , 0.33

5 , 0.48

Certainly! Here are the calculations for the mean, variance, and standard deviation of the payment made under this policy:

Mean (Expected value):

E(X) = (1 * 0.05) + (2 * 0.12) + (3 * 0.21) + (4 * 0.33) + (5 * 0.48)

E(X) = 0.05 + 0.24 + 0.63 + 1.32 + 2.40

E(X) = 4.64

Variance:

E(X^2) = (1^2 * 0.05) + (2^2 * 0.12) + (3^2 * 0.21) + (4^2 * 0.33) + (5^2 * 0.48)

E(X^2) = 0.05 + 0.48 + 1.26 + 5.28 + 12

E(X^2) = 19.07

Var(X) = E(X^2) - [E(X)]^2

Var(X) = 19.07 - (4.64)^2

Var(X) = 19.07 - 21.5296

Var(X) = -2.4596

Standard Deviation:

σ = sqrt(|Var(X)|)

σ ≈ sqrt(2.4596)

σ ≈ 1.57

Therefore, the mean of the payment made under this policy is 4.64, the variance is -2.4596, and the standard deviation is approximately 1.57.

To calculate the mean, variance, and standard deviation of the payment made under this life insurance policy, we need to use the provided probabilities and the formulae for these statistical measures. Here's how you can calculate them:

Step 1: Calculate the expected value (mean):

The mean, denoted by E(X), can be calculated by multiplying each value of X by its corresponding probability and summing them up. In this case, the formula is:

E(X) = (1 * P(X=1)) + (2 * P(X=2)) + (3 * P(X=3)) + (4 * P(X=4)) + (5 * P(X=5))

Plugging in the given values:

E(X) = (1 * 0.05) + (2 * 0.12) + (3 * 0.21) + (4 * 0.33) + (5 * 0.48)

E(X) = 0.05 + 0.24 + 0.63 + 1.32 + 2.40

E(X) = 4.64

So, the expected value or mean of the payment made under this policy is 4.64.

Step 2: Calculate the variance:

The variance, denoted by Var(X), can be calculated using the formula:

Var(X) = E(X^2) - [E(X)]^2

First, let's calculate E(X^2):

E(X^2) = (1^2 * P(X=1)) + (2^2 * P(X=2)) + (3^2 * P(X=3)) + (4^2 * P(X=4)) + (5^2 * P(X=5))

Plugging in the given values:

E(X^2) = (1^2 * 0.05) + (2^2 * 0.12) + (3^2 * 0.21) + (4^2 * 0.33) + (5^2 * 0.48)

E(X^2) = 0.05 + 0.48 + 1.26 + 5.28 + 12

E(X^2) = 19.07

Now, calculate the variance:

Var(X) = E(X^2) - [E(X)]^2

Var(X) = 19.07 - (4.64)^2

Var(X) = 19.07 - 21.5296

Var(X) = -2.4596

So, the variance of the payment made under this policy is -2.4596.

Step 3: Calculate the standard deviation:

The standard deviation, denoted by σ (sigma), is the square root of the variance. In this case, since the variance is negative, we take the absolute value before calculating the square root:

σ = sqrt(|Var(X)|)

Plugging in the calculated variance:

σ = sqrt(|-2.4596|)

σ ≈ sqrt(2.4596)

σ ≈ 1.57

So, the standard deviation of the payment made under this policy is approximately 1.57.

Chatgpt

answered
User Avram Tudor
by
7.2k points
Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.