An entrepreneur in a developing country owns 10 food carts. He has ten employees
ID: 3150707 • 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(.).
1) What is the probability that cart i will generate revenue less than $30 on a particular day?
a) 1-F(1.97)
b) F(0.625)
c) 1-F(0.625)
d) F(1.97)
2) What is the probability that average revenue will be less than $30 on a particular day?
a) 1-F(1.97)
b) F(0.625)
c) 1-F(0.625)
d) F(1.97)
3)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) 64
b) 44
c) 49
d) 7
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 = 8
Thus,
z = (x - u) / s = -0.625
Hence, as this is left tailed,
P = F(-0.625) OR by symmetry, 1 - F(0.625) [ANSWER, C]
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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
As this is left tailed, by symmetry,
OPTION A: 1 - F(1.97) [ANSWER, A]
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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 = 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, B]