Question

A person's monthly bill G has an approximate normal distribution with a variance of 2500 and a mean μ which is a function of several variables: M, the number of meals for which she will have guests, B, an index for how busy she is as work, and H, a variable which takes on 1 for the "holiday" months of November, December, and January, and 0 otherwise. Assume M∼P(8),B∼U[0,1], and μ=300+10M−100B+50H Question 7 0.0/1.0 point (graded) How is G distributed for a November where M=10 and B=.5 ?

A. G∼N(250,625)
B. G∼N(400,625)
C. G−N(250,2500)
D. G−N(400,2500)

What is F[G] ?

Answer 1

For a November with M=10 and B=0.5, the monthly bill G is normally distributed with a mean of 400 and a variance of 2500, so the correct distribution is G∼N(400, 2500). F[G] is the cumulative distribution function of G.

Step-by-step explanation:

The student is asking how the monthly bill G is distributed for a November where M (the number of meals) is equal to 10, B (an index for how busy she is at work) is equal to 0.5, and it is a 'holiday' month (where H equals 1). Given that the variance of G is 2500, and the formula for the mean μ is given as μ=300+10M−100B+50H, we can calculate:

μ = 300 + 10(10) - 100(0.5) + 50(1) = 300 + 100 - 50 + 50 = 400.

So, the mean of G for a November where M=10 and B=0.5 is 400. Since the variance is given as 2500, the standard deviation σ would be the square root of the variance, which is σ = 50. Therefore, G is normally distributed with a mean (μ) of 400 and a standard deviation (σ) of 50. Hence, G∼N(400, 2500).

The distribution function F[G] refers to the cumulative distribution function (CDF) of G, which gives the probability that G will take on a value less than or equal to a certain number.