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The highway department in a mid-sized town has enough salt to handle a total of 80 inches of snowfall. The snowfall on any given day is a random variable with mean 1.5 inches and standard deviation 0.3 inches. Assuming the snowfalls on different days are independent, approximate the probability that the highway department has enough salt to last the next 50 days.

User Miroslava
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Final answer:

To approximate the probability that the highway department has enough salt to last the next 50 days, we need to use the Central Limit Theorem and the properties of the normal distribution. The mean snowfall over 50 days is 75 inches and the standard deviation is 2.12 inches. Approximating the probability using z-scores, we find that it is approximately 0.9%.

Step-by-step explanation:

To approximate the probability that the highway department has enough salt to last the next 50 days, we need to use the Central Limit Theorem. The Central Limit Theorem states that when independent random variables are added, their sum tends to be approximately normally distributed, regardless of the shape of the original distribution.

In this case, the snowfall amounts on different days are independent and we can assume the snowfall follows a normal distribution. The mean snowfall on any given day is 1.5 inches and the standard deviation is 0.3 inches.

To find the mean and standard deviation for the sum of 50 days, we use the properties of normal distributions: The mean of the sum is equal to the sum of the individual means, and the standard deviation of the sum is equal to the square root of the sum of the individual variances.

The mean of 50 days is 50 times the mean of 1 day, which is 50 * 1.5 = 75 inches. The standard deviation of 50 days is the square root of 50 times the variance of 1 day, which is sqrt(50 * 0.3^2) = 2.12 inches.

Now we can use the normal distribution to approximate the probability that the highway department has enough salt to last the next 50 days. We want to find P(X >= 80), where X is the snowfall over 50 days. Using z-scores, we calculate the z-score as (80 - 75) / 2.12 = 2.36. Consulting a standard normal distribution table, we find that the probability of a z-score greater than 2.36 is approximately 0.009, or 0.9%.

User Chazefate
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