Transcript S2 Continuous Distributions Pupil Notes

A sprinters time for 100 metres may be recorded as 10.32 seconds
but this means that his actual time was somewhere between 10.315
and 10.325 seconds.
𝑓 𝑒 =
100
0
− 0.005 ≤ 𝑒 ≤ 0.005
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
This type of distribution is known as a uniform continuous
distribution or a rectangular distribution.
The properties of a uniform distribution are:
 It takes the same value over the range in which it is defined.
 It has no mode
 It is symmetric about its mid-range value. Therefore the mean
and median will both be equal to the mid-range value of the
distribution.
If a continuous random variable, X, has a uniform distribution over
the interval (α, β), then its probability density function will be given
by
1
𝛼 ≤𝑥 ≤𝛽
𝑓 𝑥 = 𝛽−𝛼
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Median = mean = E(X) =
𝛼+𝛽
2
Var(X) =
(𝛽− 𝛼)2
12
The cumulative distribution function is given by
0
𝑥−𝛼
𝐹 𝑥 = 𝛽−𝛼
1
𝑥 ≤ 𝛼
𝛼 ≤𝑥≤𝛽
𝑥≤ 𝛽
Example 1
The continuous random variable, X, is uniformly distributed over the
interval (5, 12). Find:
(a)
E(X)
(b)
Var(X)
(c)
P(X < 10)
Example 2
A road is being repaired and temporary telephones are installed at
intervals of 1km.
A motorist breaks down on the road and can see 100 metres in either
direction, but cannot see a phone. He tosses a coin to decide
whether to walk forwards or backwards until he finds one.
(a) Show that the distance he walks is uniformly distributed over the
interval (100, 900)
(b) Show that the mean distance he walks is half a kilometre
(c) Find the probability that he has to walk less than a quarter of a
kilometre.
Normal approximation to the binomial distribution
Continuity correction
Discrete random variables take only particular values, each with its
own probability.
Continuous random variables take values over an interval, and
probabilities are defined for ranges of values rather than individual
values.
Whenever you approximate a DISCRETE variable with a
CONTINUOUS variable, you have to apply a continuity correction.
Example 3
X is a discrete variable. Y is an approximation to X but is a
continuous variable. Write down the probability you need to
calculate for Y as the approximation for each of these probabilities
for X.
(a) P(X < 15) (b) P(X > 12) (c) P(X ≤ 17) (d) P(12 < X ≤ 15)
The parameters for the normal approximation
If X ~ B(n, p) then E(X) = np and Var(X) = npq = np(1 – p)
If n is large and p is close to 0.5, so that the distribution is nearly
symmetrical, then you can use the normal distribution to
approximate the binomial.
The general rule is that the normal distribution can be used as an
approximation when both np and np(1 – p) are > 5.
X ~ B(n, p) ≈ Y ~ N(np, np(1 – p))
Example 4
If X ~ B(30, 0.4) calculate P(8 ≤ X ≤ 15).
(a) from the tables of binomial probabilities
(b) by using a normal approximation.
Example 5
An airline estimates that 7% of passengers who book seats on a flight
do not turn up to take their seat. On one flight for which there are
185 available seats the airline sells 197 tickets. Find the probability
that they will have to refuse boarding to any passengers holding a
valid ticket for that flight.
Normal approximation to the Poisson distribution
For a Poisson distribution with large λ (> 10) you would often use a
normal distribution as an approximation, particularly when the
probability of an interval is required, e.g. P(X ≥ 15) or P(6 < X < 14),
since this will be a single calculation in a continuous distribution but
will involve multiple calculations in a discrete distribution.
X ~ Po(λ) ≈ Y ~ N(λ, λ)
Example 6
If X ~ Po(10), calculate P(8 ≤ X ≤ 15)
(a) from the tables of Poisson probabilities
(b) by using a normal approximation.
Example 7
The demand for a particular spare part in a car accessory shop may
be modelled by a Poisson distribution. On average the demand per
week for that part is 5.5.
(a) The shop has 7 in stock at the start of one week. What is the
probability that they will not be able to supply everyone who asks
for that part during the week?
(b) The manager, who is going to be away for four weeks, wants to
leave sufficient stock so that there is no more than a 5%
probability of running out of any parts while he is away. How
many of this particular spare part should he have in stock when
he leaves?