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An insurance policy for a Dyson vacuum pays $400 if it fails during the first year. The payout decreases by $100 each following year until it reaches zero. If the vacuum is still working at the beginning of any given year, the probability of it failing during that year is 0.3. Calculate the standard deviation of the payout for this Dyson vacuum under this policy.

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Final answer:

The standard deviation of the payout for this Dyson vacuum under the insurance policy is approximately $325.04.

Step-by-step explanation:

To calculate the standard deviation of the payout for the Dyson vacuum under this insurance policy, we need to first determine the payout amount for each year. Given that the payout decreases by $100 each year until it reaches zero, we can create a sequence of payments starting from $400 and subtracting $100 each year. The sequence would be: $400, $300, $200, $100, $0.

Next, we need to calculate the probability of the vacuum failing during each year. Since the probability of failure at the beginning of any given year is 0.3, the probability of the vacuum not failing is 0.7.

Finally, we can use the formula for the standard deviation of a discrete random variable to calculate the standard deviation. The formula is: sqrt(sum((x - mean)^2 * p)), where x is the payout amount, mean is the expected value (sum(x * p)), and p is the probability of each payout amount.

Applying the formula, we have:

  • For the first year: (400 - 240)^2 * 0.3 + (400 - 160)^2 * 0.7 = 14400 * 0.3 + 14400 * 0.7 = 14400.
  • For the second year: (300 - 240)^2 * 0.3 + (300 - 160)^2 * 0.7 = 3600 * 0.3 + 43560 * 0.7 = 28320.
  • For the third year: (200 - 240)^2 * 0.3 + (200 - 160)^2 * 0.7 = 14400 * 0.3 + 9000 * 0.7 = 9900.
  • For the fourth year: (100 - 240)^2 * 0.3 + (100 - 160)^2 * 0.7 = 14400 * 0.3 + 43560 * 0.7 = 28320.
  • For the fifth year: (0 - 240)^2 * 0.3 + (0 - 160)^2 * 0.7 = 43200 * 0.3 + 14400 * 0.7 = 24480.

Summing up the values from each year, we have: 14400 + 28320 + 9900 + 28320 + 24480 = 105420.

Finally, taking the square root of the sum gives us the standard deviation: sqrt(105420) ≈ 325.04.

User Amarnath Harish
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