Question

(1 point) An automobile insurer has found that repair claims are Normally distributed with a mean of $870 and a standard deviation of $810. (a) Find the probability that a single claim, chosen at random, will be less than $830.

Answer 1

The probability that a single claim, chosen at random, will be less than $830 is 0.48006.

Explanation:

We are given that an automobile insurer has found that repair claims are Normally distributed with a mean of $870 and a standard deviation of $810.

Let X = repair claims

SO, X ~ N(
`\mu = 870,\sigma^(2) = 810^(2)`)

The z-score probability distribution is given by ;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = mean claims = $870

`\sigma` = standard deviation = $810

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, the probability that a single claim, chosen at random, will be less than $830 is given by = P(X < $830)

P(X < $830) = P(
`(X-\mu)/(\sigma)` <
`(830-870)/(810)` ) = P(Z < -0.05) = 1 - P(Z
`\leq` 0.05)

= 1 - 0.51994 = 0.48006

The above probability is calculated using z table by looking at value of x = 0.05 in the z table which have an area of 0.51994.

Therefore, probability that a single claim, chosen at random, will be less than $830 is 0.48006.