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Alta Comprehensive Test Paper Questions And Solutions

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Alta Comprehensive Test Paper Questions And Solutions

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ALTA COMPREHENSIVE TEST PAPER
QUESTIONS AND SOLUTIONS

◉Which are properties of a random variable? Select all that apply.
Answer: Random variables are denoted with an upper case letter.
Random variables are not represented by numerical values.
Notation for a random variable is an upper case letter. These variables
are usually represented with words, and do not have numerical values.
Recall that discrete data are data that you can count, whether it is
random variable or not.


◉A random sample of college students majoring in cinematography
were surveyed about their movie habits. One question in the survey
asked, "How many documentaries have you watched in the past month?"
The table below represents the probability density function for the
random variable X, the number of documentaries watched by
cinematography majors in the past month. Find the standard deviation of
X.
Round the final answer to two decimal places.
x
P(X = x)
4
1/5
6

,2/5
8
2/5 Answer: $\text{std=}1.50\text{ documentaries}$std=1.50
documentaries
Here's the filled out table.
x468P(X=x)152525x⋅P(X=x)45125165(x−μ)2P(X=x)1.1520.0641.024
Note that the mean (also called expected value) μ is the sum of the
entries in the third column:μ=45+125+165=6.4This value is used to
compute the values in the fourth column. For example, the first entry
is(4−6.4)2⋅15=1.152Summing the values in the fourth column gives the
variance:1.152+0.064+1.024=2.24Taking the square root of the variance
gives the standard deviation:σ=2.24−−−−√≈1.50


◉During a bowling league tournament, the number of times that teams
scored a strike every ten minutes was recorded by a scorekeeper. The
table below represents the probability density function for the random
variable X, the number of strikes every ten minutes. Find the standard
deviation of X.
Round the final answer to two decimal places.
x
P(X = x)
0
1/6
3
1/6

,4
1/3
8
1/3 Answer: $\text{std=}2.81\text{ strikes}$std=2.81 strikes
Here's the filled out table.
x0348P(X=x)16161313x⋅P(X=x)0124383(x−μ)2P(X=x)3.3750.3750.08
334.0833
Note that the mean (also called expected value) μ is the sum of the
entries in the third column:μ=12+43+83=4.5This value is used to
compute the values in the fourth column. For example, the first entry
is(0−4.5)2⋅16=3.375Summing the values in the fourth column gives the
variance:3.375+0.375+0.0833+4.0833=7.9166Taking the square root of
the variance gives the standard deviation:σ=7.9166−−−−−√≈2.81


◉A casino features a game in which a weighted coin is tossed several
times. The table shows the probability of each payout amount (assume
that the remaining probability has a payout of 0 so that the probabilities
add to 1). To the nearest dollar, what is expected payout of the game?
Payout Amount
$130
$2,800
$135,000
Probability
0.131

, 0.021
0.0003 Answer: $116\text{ dollars}$116 dollars
The table shows the probability density function where the random
variable takes on the values 130, 2800, and 135000. To find the expected
value, multiply each payout amount by its probability and round to the
nearest dollar: $130(0.131)+$2800(0.021)+$135000(0.0003)≈$116.


◉A guessing game at a casino features 50 cards labeled with the
numbers 1 through 50. Four cards will be drawn without replacement
and each player will guess the card numbers. The probability of each
payout amount is shown in the table (assume that the remaining
probability has a payout of 0 so that the probabilities add to 1). To the
nearest dollar, what is the expected payout of the game?
Payout Amount
$100
$1,400
$180,000
Probability
0.106
0.023
0.0002 Answer: $79$79
The table shows the probability density function where the random
variable takes on the values 100, 1400, and 180000. To find the expected
value, multiply each payout amount by its probability and round to the
nearest dollar: $100(0.106)+$1400(0.023)+$180000(0.0002)=$79.

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