Quiz 5 – Sections 4.4 – 5.2
Stat 216
Name ________________________
Class ID #: _______
1. TRUE or FALSE. As the sample size increases, the sample mean gets closer and closer to the
population mean. (1 pt)
2. TRUE or FALSE. Sample means are more spread out than individual observations. (1 pt)
3. You buy a hot stock for $1000. The stock either gains 30% or loses 25% each day, each with
probability of 0.5. Its returns on consecutive days are independent of each other. You will sell the
stock after two days.
(a) Make a probability distribution showing the possible values of the stock after two days and the
probability for each value. (2 pts)
S = {GG, GL, LG, LL}
For GG: X = (1000 x 1.3) x 1.3 = 1690
For GL: X = (1000 x 1.3) x 0.75 = 975
For LG: X = (1000 x 0.75) x 1.3 = 975
For LL: X = (1000 x 0.75) x 0.75 = 562.50
P(GG) = (0.5)(0.5) = 0.25
P(GL) = (0.5)(0.5) = 0.25
P(LG) = (0.5)(0.5) = 0.25
P(LL) = (0.5)(0.5) = 0.25
X = stock value
Probability
1690
1/4
975
1/2
P(X=1690) = P(GG) = 0.25
P(X=975) = P(GL or LG) = P(GL) +
P(LG) = 0.25 + 0.25 = 0.5
P(X=562.50) = P(LL) = 0.25
562.50
1/4
(b) What is the mean value of the stock after two days? (1 pt)
µx = 1690(1/4) + 975(1/2) + 562.50(1/4) = 1050.63
(c) Is it a good idea to pay $1000 for this stock? Why or why not? (1 pt)
No, even though µx > 1000, because you are only doing this once and P(X < 1000) = 3/4.
However, if you were to repeat this process of buying and selling such a stock after two days over
and over again, you could expect to gain approximately $50 per time. This would eventually be
profitable (and therefore worth the $1000 price), but only if the action was to be repeated many
times over.
4. An enthusiastic weatherman states that the annual precipitation X in Illinois is normally distributed
with mean of 33 inches and standard deviation of 4 inches.
(a) Find the probability that, in a randomly selected year, more than 40 in will fall in Illinois? (1 pt)
X ~ N(33, 4)
z = (40-33)/4 = 1.75
P(X > 40) = P(Z > 1.75) = 1 – P(Z < 1.75) = 1 – 0.9599 = 0.0401
(b) Suppose we take a simple random sample of 50 years, then calculate the sample mean
precipitation. What is the sampling distribution of X ? (1 pt)
X ~ N (33, 4/ 50 )
or
X ~ N (33, 0.5657)
(c) Find the probability that the sample mean precipitation is between 30 and 40? (2 pts)
!
!
z = (30-33)/0.5657
= -5.3
z = (40-33)/0.5657 = 12.37
P(30 < X < 40) = P(-5.3 < Z < 12.37) = P(Z < 12.37) – P(Z < -5.3) ! 1 – 0 = 1
***Bonus*** State the Central Limit Theorem = If the X distribution is not normal, then the
X distribution is approximately normal X ~ N(µ, "/ n ) if n > 30.
!