Natalie and Michelle are roommates that both work as waitresses in two different

Question

Natalie and Michelle are roommates that both work as waitresses in two different restaurants. The amount of money that Michelle earns in a week is a random variable with a mean of $50o and a standard deviation of $75. The amount of money that Natalie earns in a week is a random variable with a mean of $600 and a standard deviation of $100. The two of them work in different parts of Austin, so we will assume that the two ladies' earnings are independent of one another. We will also assume that the distributions for each ladies' weekly earnings are approximately normally distributed. 3. a) What are the mean and standard deviation for Michelle and Natalie's combined earnings (M N) for one week? b) What is the probability that Michelle and Natalie's combined weekly income exceeds $1300? (In other words, what is the probability that the variable "M N" is at least $130000?) c) What are the mean and standard deviation for the difference in (N - M), Natalie's earnings and Michelles earnings for one week? d) What is the probability that Natalie's weekly income is at least $240.00 more than Michelle's weekly income? (In other words, what is the probability that the difference "N - M" is at least $240.oo?)

Explanation / Answer

Let M be the Michelle's earning and N be the Natalie's earning.

Given that M~ N(mu = $500, sd = $75)

and N ~ N(mu = $600, sd = $ 100)

M and N are independent.

a) Here M+N follows also normal distribution with mean (mu) = $500+$600 = $1100

Var = 75^2 + 100^2 = 15625

sigma = sqrt(15625) = 125

So M+N ~ N(mu = $1100, sd = $125)

b) In this part we have to find P(M+N >= 1300)

Now we have to find z-score for M+N= 1300

z-score is defined as,

z = (x - mean) / sd

z = (1300 - 1100) / 125 = 1.6

Now we have to find P(Z >= 1.6)

This probability we can find in excel.

syntax :

=1 - NORMSDIST(z)

where z is z-score.

P(Z >= 1.6) = 0.0548

c) Now we have to find distribution of N-M.

The difference between N and M also follows normal distribution with mean = 600-500 = 100 and sd = 125.

Therefore, N - M ~ N(mu = 100, sd = 125).

d) Here we have to find P(N-M >= 240)

z-score for N-M = 240 is,

z= (240 - 100) / 125 = 1.12

Now we have to find P(Z > = 1.12)

P(Z >= 1.12) = 0.1314

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