An entrepreneur in a developing country owns 10 food carts. He has ten employees
ID: 3151282 • Letter: A
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.
In this question the cumulative distribution function of the standard normal random variable is denoted by F(.).
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What is the probability that cart i will generate revenue less than $30 on a particular day?
A. F(1.97)
B. 1-F(0.625)
C. F(0.625)
D. 1-F(1.97)
What is the probability that average revenue will be less than $30 on a particular day?
A. F(1.97)
B. 1-F(0.625)
C. F(0.625)
D. 1-F(1.97)
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?
A. 44
B. 7
C. 49
D. 64
Explanation / Answer
1.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 30
u = mean = 35
s = standard deviation = sqrt(64) = 8
Thus,
z = (x - u) / s = -0.625
By symmetry, the left tailed area is
P(z<-0.625) = 1 - F(0.625) [ANSWER, B]
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2.
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 30
u = mean = 35
n = sample size = 10
s = standard deviation = 8
Thus,
z = (x - u) * sqrt(n) / s = -1.976423538
By symmetry, the left tailed area is
P(z<-1.97) = 1 - F(1.97) [ANSWER, D]
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3.
That means a margin of error of +/- 2 from the mean.
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = (1-0.90)/2 = 0.05
Using a table/technology,
z(alpha/2) = 1.644853627
Also,
s = sample standard deviation = 8
E = margin of error = 2
Thus,
n = 43.28869527
Rounding up,
n = 44 [ANSWER, A]