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S2 Notes (Edexcel)



Various Things to Remember for S2

Necessary conditions for using the Binomial Distribution Binomial Distribution can be used under the following conditions: • The number of trials is fixed • The trials are independent of each other • There are two outcomes – success and failure • The probability of success in each trial is constant. Necessary conditions for using the Poisson Distribution • The events occur at random • The events are independent of each other • The average rate at which the events occur remains constant • Two or more events cannot occur at the same time

Poisson as an Approximation to Binomial Poisson Distribution can be used an approximation for Binomial when • n is large (n > 50 ) and p is sufficiently small (np < 5 ). If the probability of success p is close to 1 then consider the probability of failure q and use Poisson as a model for number of failures. June 2007 Q5 Write down the conditions under which the Poisson distribution may be used as an approximation to the Binomial distribution.



A reminder from S1 about skewness Skewness A distribution that is not symmetric is said to be skewed.

Two tests for skewness

Positive skew if ( ) ( )3 2 2 1 0Q Q Q Q− − − > and if Mean > Median > Mode

Negative skew if ( ) ( )3 2 2 1 0Q Q Q Q− − − < and if Mean < Median < Mode

A good way to remember the condition for skewness involving mean, median and mode. Write down the three averages in alphabetical order, that is

Mean…Median…Mode

For positive skew fill the blank with a “>” sign to give

Mean > Median > Mode For negative skew fill the blank with a “<” sign to give

Mean < Median < Mode

Mean Median

Mode

Negative Skew

Mean Median Mode

Positive Skew If the distribution has a long ‘tail’ of high values then it is said to be positively skewed. In this case the median is less than the mean.

If the distribution has a long ‘tail’ of low values then it is said to be negatively skewed. In this case the median is greater than the mean.



Binomial approximated by Normal So if ~ ( , )X B n p then the approximation is ( , )X N np npq≈ where 1q p= − . This can be used provided that 10n > and provided that 5np > and 5nq > .

That double condition 5np > and 5nq > means that p must be close to 12

for values of n that are

not very large. NB : Since the Normal distribution is symmetrical the normal distribution fits the Binomial best

when 12

p = .

June 2007, Q7: Write down two conditions for X ~ Bin(n, p) to be approximated by a normal distribution Y ~ N (µ, σ 2).

Poisson approximated by Normal If o~ ( )X P λ then the approximation ~ ( , )X N λ λ can be used provided that 10λ > . Normal or Poisson Approximation to Binomial? What to do if 5 10np≤ ≤ is slightly confusing. As a general rule, if 0.1p ≤ use Poisson but if

0.1p > use Normal.



Definitions Census: A census is when every member of the population is investigated. January 2008, Q1 Critical region: The critical region is the range of values that would lead to the rejection of 0H . January 2008 Q7 Hypothesis test: A hypothesis test is a mathematical procedure to examine a value of the population parameter proposed by the null hypothesis compared with an alternative hypothesis. January 2008, Q7 Population: A population is collection of all items. June 2010 Sampling distribution: All possible samples are chosen from a population; the values of a statistic and the associated probabilities is a sampling distribution. June 2005, Q4 Sampling frame: A sampling frame is a list of all sampling units or all the population. June 2005, Q4 Sampling unit: A sampling unit is an individual member or element of the population or sampling frame. June 2005, Q4 Significance Level: The probability of incorrectly rejecting 0H or Probability of rejecting 0H when

0H is true. January 2012, Q7a Statistic: A statistic is a function of the sample which contains no unknown parameters. June 2010



Binomial Distribution

The binomial distributions. Candidates will be expected to use these distributions to model a real-world situation and to comment critically on their appropriateness. Cumulative probabilities by calculation or by reference to tables. The Binomial distribution is used when there are a set number of trials, each of which can have two outcomes. If there are n trials, if the probability of each trial being a ‘success’ is p and if X is the number of successes then X is said to be binomially distributed with parameters n and p. In this case write ),(~ pnBX . RECAP FROM C2 If n is a positive integer then ( ) ( )! 1 2 .... 2 1n n n n= × − × − × × × . In how many ways can the letters p, q, r, s, t be arranged? The first could be any one of 5 letters, the second any one of 4 and so on. Hence the letters p, q, r, s, t can be arranged in 5! ways. If p, q and r are replaced with the letter a then this will clearly reduce the number of arrangements For example there were six ways in which st could be the last two letters, that is pqrst, prqst, qprst, qrpst, rpqst and rqpst but now there is just one, that is aaast. So the total number of arrangements is reduced by a factor of 3! (since p, q and r can be arranged in 3! ways).

Hence the letters a a a s t can be arranged in 5!3!

ways.

If s and t are replaced with the letter b then this will further reduce the number of arrangements For example there were two ways in which aaa could be the first three letters, that is aaast, aaats, but now there is just one, that is aaabb. So the total number of arrangements is reduced by a factor of 2! (since s and t can be arranged in 2! ways).

Hence the letters a a a b b can be arranged in 5!3!2!

ways.

In general consider this arrangement of letters a a a a a a a a a a a b b b b b b

These can be arranged in ( )

!! !

nr n r−

ways. The notation nr

or nrC is used to represent

( )!

! !n

r n r−.

r n r−



And so

( )!

1 1! 1 !n n n

n

= = − ,

( )( )1!

2 2! 2 ! 1 2n nn n

n−

= = − × and

( )( )( )1 2!

3 3! 3 ! 1 2 3n n nn n

n− −

= = − × ×

and in general

( )( )( ) ( )1 2 .... 1!

! ! 1 2 3 ...n n n n rn n

r r n r r− − − +

= = − × × × ×

NB : 0! 1= and so ! 10 0! !n n

n

= =

and ! 1!0!

n nn n

= =

Example A die is thrown 4 times and X is the number of sixes thrown. X can take the values 0, 1, 2, 3, or 4. Suppose s represents getting a six and n represents not getting a six. The only way of getting four sixes can be represented as s s s s. Since the probability of any single throw being a six is 5

6 .

( ) ( )456P s s s s = and so ( )45

6( 0)P X = = . The ways of getting three sixes can one non-six be represented as n s s s s n s s s s n s s s s n This fits in with the previous page, that there are 4

1C ways of arranging the letters s s s n.

The probability of each of these events happening is ( )3516 6× .

So ( )34 511 6 6( 1)P X C= = × × .

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