An entrepreneur in a developing country owns 10 food carts. He has ten employees

Question

An entrepreneur in a developing country owns 10 food carts. He has ten employees to work with these food carts. Let Xi be a random variable representing revenue from cart i (on a particular day), i = 1,..., 10. Xi is approximately normally distributed with mean $35, and variance 64 (squared dollars). Revenues of the different carts are independent. How many carts would the entrepreneur have to own in order for the probability to be at least 0.90 that average revenue on a particular day will be between $33 and $37? The cumulative distribution function of a normal random variable is denoted by F(.) in this exercise.

Explanation / Answer

For the middle 0.90 of the values, the tails will have an area of (1-0.90)/2 = 0.05. Hence, the corresponding z scores are, by table/technology,

z = -1.644853627, 1.644853627

Thus, for the upper endpoint, x = 37, as

z = (x-u)*sqrt(n)/sigma

Then, as sigma = sqrt(64) = 8,

n = [z*sigma/(x-u)]^2 = (1.644853627*8/(37-35))^2 = 43.28869527

Rounding up,

n = 44 [ANSWER]

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