MA22S6 Homework 3 Solutions • Question 1

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MA22S6 Homework 3 Solutions
• Question 1
The number of annual hurricanes, X, can be expressed via a Poisson distribution with
rate parameter of 6, i.e., X ∼ Pois(6). The probability of k in a year is P(X = k) =
6k −6
k! e .
The probability of less than four is
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
=
60 −6 61 −6 62 −6 63 −6
e + e + e + e
0!
1!
2!
3!
= 0.00247 + 0.0148 + 0.044 + 0.089 ≈ 0.15.
The probability of between five and eight in the year is
P(5 ≤ X ≤ 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
=
65 −6 66 −6 67 −6 68 −6
e + e + e + e
5!
6!
7!
8!
= 0.16 + 0.16 + 0.1376 + 0.1032 ≈ 0.56.
We can also consider the number of earthquakes as X ∼ Bin(30, 0.2), giving P(X = k) =
(30
)0.2k 0.830−k ). This leads to
k
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.001 + 0.009 + 0.034 + 0.078 = 0.122
P(5 ≤ X ≤ 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
= 0.172 + 0.179 + 0.153 + 0.11 ≈ 0.614
We see the figures from both distributions are similar.
• Question 2
Let X be the number of defective items per set of ten. The probability of at most one
defective item is P(X ≤ 1) = P(X = 0) + P(X = 1). If X is Binomia with probability
1
0.1 of an item being defective, i.e., X ∼ Bin(10, 0.1),, the probability of k defectives is
)(0.1)k (0.9)10−k , yielding
P(X = k) = (10
k
P(X ≥ 1) = P(X = 0) + P(X = 1)
10
10
= ( )(0.1)0 (0.9)10 + ( )(0.1)1 (0.9)9
0
1
= 0.3486 + 0.3874 ≈ 0.736.
We can think of the occurence of defectives as Poisson with a rate of λ = 1 per 10 items,
i.e., X ∼ Pois(1). The probability of k items being defective is P(X = k) =
1k −1
k! e ,
giving
P(X ≥ 1) = P(X = 0) + P(X = 1)
=
10 −1 11 −1
e + e
0!
1!
= 0.367 + 0.367 ≈ 0.734.
These two figures are very similar.
• Question 3
X ∼ Bin(8, 0.2) implies the probability of k successes in 8 trials is P(X = k) = (k8)(0.2)k (0.8)8−k .
Calculating P(X = k) for k = 0, 1, . . . , 8 and plotting against k gives Writing λ = np =
8(0.2) = 1.6 we can also say X ∼ Pois(1.6) giving We see both plots are very similar.
2
• Question 4
The cumulative density function of the uniform distribution on [a, b] (the probability
of a value less than k) is P(X < k) =
k−a
b−a .
The probability of an outcome exceeded a
particular value is P(X > k) = 1 − P(X < k) = 1 −
P(X < 99) =
99−98.5
99.5−98.5
= 0.5 and P(X > 99.4) =
k−a
b−a
99.5−99.4
99.5−98.5
=
b−k
b−a .
From these identities we see
= 0.1. The probability of a value of
99 is zero, as the uniform distribution is continuous (i.e., can take an infinite amount of
values).
3
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