Question

the table shows claims and their probabilities for an insurance company. use the table to answer parts a through c.

amount of claim (to the nearest $50,000)	probability
$50,000	0.21
$100,000	0.07
$150,000	0.05
$200,000	0.01
$250,000	0.01

Explanation:

Since the specific parts (a through c) of the question are not provided, I'll assume common questions related to this probability distribution table, such as finding the expected value (mean) of the claim amount. Here's the solution for that:

Step1: Recall the formula for expected value

The expected value \( E(X) \) of a discrete random variable \( X \) is given by \( E(X)=\sum_{i} x_i P(x_i) \), where \( x_i \) are the possible values of \( X \) and \( P(x_i) \) are their corresponding probabilities.

Step2: Identify \( x_i \) and \( P(x_i) \)

From the table:

• \( x_1 = 0 \), \( P(x_1)=0.85 \)
• \( x_2 = 50000 \), \( P(x_2)=0.21 \)
• \( x_3 = 100000 \), \( P(x_3)=0.07 \)
• \( x_4 = 150000 \), \( P(x_4)=0.05 \)
• \( x_5 = 200000 \), \( P(x_5)=0.01 \)
• \( x_6 = 250000 \), \( P(x_6)=0.01 \)

Step3: Calculate each term \( x_i P(x_i) \)

• For \( x_1 \): \( 0 \times 0.85 = 0 \)
• For \( x_2 \): \( 50000 \times 0.21 = 10500 \)
• For \( x_3 \): \( 100000 \times 0.07 = 7000 \)
• For \( x_4 \): \( 150000 \times 0.05 = 7500 \)
• For \( x_5 \): \( 200000 \times 0.01 = 2000 \)
• For \( x_6 \): \( 250000 \times 0.01 = 2500 \)

Step4: Sum up all the terms

\( E(X)=0 + 10500+7000 + 7500+2000 + 2500\)
\( E(X)=10500+7000=17500\); \( 17500 + 7500 = 25000\); \( 25000+2000 = 27000\); \( 27000+2500 = 29500\)

Answer:

The expected claim amount is \(\$29500\) (if the question was to find the expected value). If your question was different (like finding the probability of a claim exceeding a certain amount, or variance, etc.), please provide more details.