Download RZC-Chp6-ProbabilityStats-Worksheet2

Probability/Stats Worksheet 2 – Binomial/Geometric/Poisson/Gaussian
Distributions
1. For the following, provide (i) how you would define a random variable 𝑋 with the
appropriate distribution and (ii) calculate the required probability.
a. Example: Throwing 0 Heads in 4 throws of a fair coin.
i. 𝑋~𝐵(4, 0.5)
ii. 𝑃(𝑋 = 0) = 0.54 = 0.0625
b. Throwing 2 Heads in 4 throws of a biased coin, with probability of Heads 0.4.
c. Throwing a biased coin, with probability of Heads 0.7, until you see a Heads on the
5th throw.
d. Receiving 5 visits to your website in the next minute, given that you see 4 visits on
average.
2. By finding a Z-Table (just Google it!) determine the probability of these particular IQ ranges,
given that the mean IQ is 100 and the Standard Deviation is 15.
a. An IQ less than 100.
b. An IQ less than 115.
c. An IQ less than 70 (your Z-Table might only have positive values of 𝑧; if so, think
about the symmetry of the Normal Distribution).
d. An IQ greater than 142.
e. An IQ greater than 87.
f. An IQ between 90 and 120.
3. [Source: University interview] In an experiment to investigate animal behaviour rats have to
make a choice between 4 doors of different colours. If they make the right choice they find
food; if they make the wrong choice they get an electric shock. If an incorrect choice is
made, the animal returns to its starting point and tries again, and this continues until a
correct choice is made.
a. If 𝑝(𝑥) is the probability that a correct choice is made at the 𝑥th attempt, find 𝑝(𝑥)
if:
i. each door is equally likely to be chosen at each trial, and all trials are
mutually independent;
ii. at each trial the rat chooses with equal probability between the doors which
have not been tried previously, no choice ever being repeated;
iii. the rat never chooses the same door on two successive trials but otherwise
chooses at random.
b. Which strategy would be the best one for the rat to adopt?
4. [Source: STEP] The number of texts that George receives on his mobile phone can be
modelled by a Poisson random variable with mean 𝜆 texts per hour.
Given that the probability George waits between 1 and 2 hours in the morning before he
receives his first text is 𝑝, show that:
𝑝𝑒 2𝜆 − 𝑒 𝜆 + 1 = 0
Hint: the volcano example in the slides might be helpful.
𝑑
5. (Difficult!) By noting that 𝑑𝑝 [−(1 − 𝑝)𝑥 ] = 𝑥(1 − 𝑝)𝑥−1 , prove that if 𝑋~𝐺𝑒𝑜𝑚(𝑝) (i.e. the
1
𝑝
geometric distribution with probability of success 𝑝), then 𝐸[𝑋] = .
www.drfrostmaths.com/rzc
ANSWERS
1.
Answers
b. 𝑋~𝐵(4, 0.4)
c. 𝑋~𝐺𝑒𝑜𝑚(0.7)
d.
𝑋~𝑃𝑜𝑖𝑠𝑠(4)
𝑃(𝑋 = 2) = 0.3456
𝑃(𝑋 = 5) = 0.34 × 0.7 = 0.00567
𝑃(𝑋 = 5) =
45
5!
𝑒 −4 = 0.156
2. Answers:
a. An IQ less than 100: 𝑝(𝑧 ≤ 0) = 0.5
b. An IQ less than 115: 𝑝(𝑧 ≤ 1) = 0.8413
c. An IQ less than 70: 𝑝(𝑧 ≤ −2) = 1 − 𝑝(𝑧 ≤ 2) = 0.0228
142−100
d. An IQ greater than 142: 142 is 15 = 2.8𝜎s above the mean. 𝑝(𝑧 > 2.8) = 1 −
𝑝(𝑧 ≤ 2.8) = 1 − 0.9974 = 0.0026
e. An IQ greater than 87: By symmetry, this is the same as having an IQ less than 113.
113−100
113 is
= 0.87𝜎s away from the mean. 𝑝(𝑧 < 0.87) = 0.8078
15
f. An IQ between 90 and 120: The probability of an IQ less than 120 is 0.9082.
Therefore the probability of an IQ between 100 and 120 is 0.9082-0.5=0.4082. The
probability of an IQ greater than 90 is the same as the probability of an IQ less than
110, which is 0.7454. So the probability of an IQ between 90 and 100 is 0.74540.5=0.2454.
Therefore the probability of an IQ between 90 and 120 is 0.2454 + 0.4082 =
0.6536.
3. Answers
a. Answers:
3 𝑥−1 1
,
𝑥=
4
1
𝑥 = 1,2,3,4
{4
i. 𝑝(𝑥) = (4)
ii. 𝑝(𝑥) =
0
iii. 𝑝(𝑥) = {1
1
4
1,2, …
This is a geometric distribution.
𝑥>4
𝑥 = 1,2
2 𝑥−2
( )
4 3
𝑥>2
b. The most obvious thing to find is the expected value of each distribution, since this
gives the expected number of times the rat will have to choose a door (obviously,
1
the less the better!). Recall that if 𝑋~𝐺𝑒𝑜𝑚(𝑝), then 𝐸[𝑋] = 𝑝. Since the first
1
strategy is 𝐺𝑒𝑜𝑚 (4), the expected number of doors is 4. For the second strategy,
1
𝐸[𝑋] = 4 (1 + 2 + 3 + 4) = 2.5. For the last strategy, via some appropriate
manipulation we get 𝐸[𝑋] = 3.25. The second strategy is therefore the best.
Alternatively, you might look at the cumulative distribution function of each, but this
would be difficult to calculate for all but the first strategy.
4. Let 𝑋 be the number of texts George receives in the first hour of the morning and 𝑌 be the
number he receives in the second hour.
Then 𝑋~𝑃𝑜(𝜆) and 𝑌~𝑃𝑜(𝜆), with 𝑋 and 𝑌 independent random variables.
This we have
𝑃(𝐺𝑒𝑜𝑟𝑔𝑒 𝑤𝑎𝑖𝑡𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 1 𝑎𝑛𝑑 2 ℎ𝑜𝑢𝑟𝑠 𝑓𝑜𝑟 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑥𝑡) = 𝑃(𝑋 = 0 𝑎𝑛𝑑 𝑌 > 0)
= 𝑃(𝑋 = 0). 𝑃(𝑌 > 0)
= 𝑒 −𝜆 . (1 − 𝑒 −𝜆 )
=𝑝
www.drfrostmaths.com/rzc
Then 𝑒 −𝜆 − 𝑒 −2𝜆 = 𝑝
Multiplying this last equation by 𝑒 2𝜆 gives 𝑒 𝜆 − 1 = 𝑝𝑒 2𝜆 , and a straightforward
rearrangement yields our desired equation.
Note that there is a distribution known as the exponential distribution which would directly
allow us to find the probability of waiting a particular amount of time.
𝑥−1
5. 𝐸[𝑋] = ∑∞
𝑝
𝑥=0 𝑥(1 − 𝑝)
∞
= 𝑝∑
𝑥(1 − 𝑝)𝑥−1
𝑥=0
∞
𝑑
[−(1 − 𝑝)𝑥 ]
𝑥=0 𝑑𝑝
∞
𝑑
= −𝑝 [∑ (1 − 𝑝)𝑥 ]
𝑑𝑝
𝑥=0
We were able to move the differential function outside the summation, given that when we
= 𝑝∑
𝑑
differentiate a sum, we differentiate each of the terms in turn, i.e. 𝑑𝑝 [𝑓(𝑝) + 𝑔(𝑝)] =
𝑑
[𝑓(𝑝)]
𝑑𝑝
𝑑
+ 𝑑𝑝 [𝑔(𝑝)].
We now conveniently have an infinite Geometry Series, where 𝑎 = 1 − 𝑝 and 𝑟 = 1 − 𝑝.
𝑎
Thus using 𝑆∞ = 1−𝑟, we get
1−𝑝
𝑝
= 𝑝−1 − 1. So continuing:
𝑑 −1
(𝑝 − 1)
𝑑𝑝
= −𝑝(−𝑝−2 )
1
=
𝑝
= −𝑝
www.drfrostmaths.com/rzc