Discrete Random Variables, normal distribution

Discrete Random Variables, normal distribution

HideShow resource information
Preview of Discrete Random Variables, normal distribution

First 317 words of the document:

S1 Revision Notes: Discrete Random Variables
Recap: In a statistical experiment the outcome is unknown in advance. The outcome of the experiment is
called a random variable. If the experiment can result in only a fixed number of outcomes, it is called a
discrete random variable.
The outcomes of the experiment and the probabilities form a probability distribution. They can be shown in
a table:
A discrete random variable has certain properties:
The sum of all the probabilities is 1, i.e.
Each probability is between 0 and 1, i.e.
There are 2 key formulae:
E(X) = (i.e. times the top and bottom rows together and add).
Var(X) = E(X²) ² where = E(X)
and E(X²) =
Example:
A shop sells gift vouchers valued at £1, £2, £4, £10 or £20. The value in £ of a gift voucher sold may be
regarded as a random variable, X, with the following distribution:
x 1 2 4 10 20
P(X = x) 0.20 0.40 0.22 0.11 0.07
a) Find the mean and the standard deviation of X.
b) What is the probability that the value of the next voucher sold is less than £4.
The shop is considering whether to discontinue selling £1 and £2 vouchers. If they did this a proportion p
of customers who presently buy £1 and £2 vouchers would then buy a £4 voucher. Other such customers
would not buy a voucher. As a result the value, in £, of sales would be a random variable Y with the
following distribution.
y 0 4 10 20
P(Y = y) 0.6(1 ­ p) 0.22 + 0.6p 0.11 0.07
c) Find the mean of Y in terms of p.
d) A survey suggests that the value of p would be between 0.5 and 0.7. Using this information,
advise the shop on the likely effect on takings if she decides to discontinue selling £1 and £2
vouchers.
1
Created on 12/05/2005 10:12 AM by A. Duncombe

Other pages in this set

Page 2

Preview of page 2

Here's a taster:

Solution
a) E(X) = (1 × 0.20) + (2 × 0.40) + (4 × 0.22) + (10 × 0.11) + (20 × 0.07) = £4.38
E(X²) = (1² × 0.20) + (2² × 0.40) + (4² × 0.22) + (10 ² × 0.11) + (20² × 0.07)
= 44.32
So Var(X) = 44.32 ­ 4.38² = 25.1356
So SD = £5.01.
b) P(X < 4) = P(X = 1 or 2) = 0.20 + 0.40 = 0.…read more

Comments

No comments have yet been made

Similar Statistics resources:

See all Statistics resources »See all resources »