Chapter 2

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Chapter 2
Verify that the following functions are probability mass functions, and determine the
requested probabilities.
1.
a)
b)
c)
d)
𝑷(𝑿 ≤ 𝟐)
𝑷(𝑿 > −𝟐)
𝑷(−𝟏 ≤ 𝑿 ≤ 𝟏)
𝑷(𝑿 ≤ −𝟏 𝒐𝒓 𝑿 = 𝟐)
Solution:
-
All probabilities are greater than or equal to zero and sum to one.
a) 𝑃(𝑋 ≤ 2) = 1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1
b) 𝑃(𝑋 > −2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8
c) 𝑃(−1 ≤ 𝑋 ≤ 1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4
d) 𝑃(𝑋 ≤ −1 π‘œπ‘Ÿ 𝑋 = 2) = 1/8 + 2/8 +1/8 = 4/8 =1/2
2.
𝒇(𝒙) =
πŸπ’™+𝟏
πŸπŸ“
,
x = 0,1,2,3,4
a) 𝑷(𝑿 = πŸ’)
b) 𝑷(𝑿 ≤ 𝟏)
c) 𝑷(𝟐 ≤ 𝑿 < πŸ’)
d) 𝑷(𝑿 > −𝟏𝟎)
Solution:
-
Probabilities are nonnegative and sum to one
a) 𝑃(𝑋 = 4) = 9/25
b) 𝑃(𝑋 ≤ 1) = 1/25 + 3/25 = 4/25
c) 𝑃(2 ≤ 𝑋 < 4) = 5/25 + 7/25 = 12/25
d) 𝑃(𝑋 > −10) = 1
3. The thickness of wood panelling (in inches) that a customer orders is a random
variable with the following cumulative distribution function:
Determine the following probabilities
a) 𝑷(𝑿 ≤ 𝟏/πŸ–)
b) 𝑷(𝑿 ≤ 𝟏/πŸ’)
c) 𝑷(𝑿 ≤ πŸ“/πŸπŸ”)
d) 𝑷(𝑿 > 𝟏/πŸ’)
e) 𝑷(𝑿 ≤ 𝟏/𝟐)
Solution:
-
The sum of the probabilities is 1 and all probabilities are greater than or equal to zero.
-
P.m.f: f (1/8) = 0.2, f (1/4) = 0.7, f (3/8) = 0.1
a) 𝑃(𝑋 ≤ 1/8) = 0
b) 𝑃(𝑋 ≤ 1/4) = 0.9
c) 𝑃(𝑋 ≤ 5/16) = 0.9
d) 𝑃(𝑋 > 1/4) = 0.1
e) 𝑃(𝑋 ≤ 1/2) = 1
4. Suppose that 𝒇(𝒙) = 𝒆−(𝒙−πŸ’) 𝒇𝒐𝒓 𝒙 > πŸ’. Determine the following probabilities:
a) 𝑷(𝑿 > 𝟏)
b) 𝑷(𝟐 ≤ 𝑿 < πŸ“)
c) 𝑷(𝑿 > πŸ“)
d) 𝑷(πŸ– < 𝑿 < 𝟏𝟐)
e) Determine x, such that 𝑷(𝑿 < 𝒙) = 𝟎. πŸ—
Solution:
5. Suppose the cumulative distribution function of the random variable X is:
Determine the following:
a) 𝑷(𝑿 < 𝟏. πŸ–)
b) 𝑷(𝑿 > −𝟏. πŸ“)
c) 𝑷(𝑿 < −𝟐)
d) 𝑷(−𝟏 < 𝑿 < 𝟏)
Solution:
6. Suppose the probability density function of the length of computer cables is f(x) = 0.1
from 1200 to 1210 millimetres.
a) Determine the mean and standard deviation of the cable length.
b) If the length specifications are 1195 < x < 1205 millimetres, what proportion of
cables are within specifications?
Solution:
7. From a box containing 4 black balls and 2 green balls, 3 balls are drawn in
succession, each ball being replaced in the box before the next draw is made. Find
the probability distribution for the number of green balls.
Solution:
Denote by X the number of green balls in the three draws. Let G and B stand for the colors of
green and black, respectively.
The probability mass function for X is then:
8. Suppose a special type of small data processing firm is so specialized that some have
difficulty making a profit in their first year of operation. The p.d.f that characterizes
the proportion Y that make a profit is given by:
𝒇(π’š) = π’Œπ’šπŸ’ (𝟏 − π’š)πŸ‘ ,
𝟎 ≤ π’š ≤ 𝟏 and 𝒇(π’š) = 𝟎, π’†π’π’”π’†π’˜π’‰π’†π’“π’†
a) What is the value of k that renders the above a valid density function?
b) Find the probability that at most 50% of the firms make a profit in the first
year.
c) Find the probability that at least 80% of the firms make a profit in the first
year.
Solution:
1
a) Using integral by parts and setting π‘˜ ∫0 𝑦 4 (1 − 𝑦)3 𝑑𝑦 = 1, we obtain k = 280.
b) For 0 ≤ 𝑦 ≤ 1, 𝐹(𝑦) = 56𝑦 5 (1 − π‘Œ)3 + 28𝑦 6 (1 − 𝑦)2 + 8𝑦 7 (1 − 𝑦) + 𝑦 8
So, 𝑃(π‘Œ ≤ 5) = 0.3633
c) Using C.d.f in (b), 𝑃(π‘Œ > 0.8) = 0.0563
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