Answer:
0.0287 = 2.87% probability the total of the 100 claims exceeds 1,530,000.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
n instances of a normal variable:
For n instances of a normal variable, the mean is [tex]n\mu[/tex] and the standard deviation is [tex]s = \sigma\sqrt{n}[/tex]
Sum of normal variables:
When two normal variables are added, the mean is the sum of the means, while the standard deviation is the square root of the sum of the variances.
Group A follow a normal distribution with mean 10,000 and standard deviation 1,000. 50 claims of group A.
This means that:
[tex]\mu_A = 10000*50 = 500000[/tex]
[tex]s_A = 1000\sqrt{50} = 7071[/tex]
Group B follow a normal distribution with mean 20,000 and standard deviation 2,000. 50 claims of group B.
This means that:
[tex]\mu_B = 20000*50 = 1000000[/tex]
[tex]s_B = 2000\sqrt{50} = 14142[/tex]
Distribution of the total of the 100 claims:
[tex]\mu = \mu_A + \mu_B = 500000 + 1000000 = 1500000[/tex]
[tex]s = \sqrt{s_A^2+s_B^2} = \sqrt{7071^2+14142^2} = 15811[/tex]
Find the probability the total of the 100 claims exceeds 1,530,000.
This is 1 subtracted by the p-value of Z when X = 1530000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1530000 - 1500000}{15811}[/tex]
[tex]Z = 1.9[/tex]
[tex]Z = 1.9[/tex] has a p-value of 0.9713
1 - 0.9713 = 0.0287
0.0287 = 2.87% probability the total of the 100 claims exceeds 1,530,000.