S2 Statistics 2

Example:Find the number of hands of 4 cards which can be dealt from a pack of 10.

Solution:In a hand of cards the order does not matter, so this is just the number of combinations of 4 from 10, or 10 choose 4

notice 4 numbers on top of the fraction

Example:10 dice are rolled. Find the probability that there are 4 sixes.

Solution:If X is the number of sixes then  X ∼  B10,  16

We could have  6,6,6,6,×,×,×,×,×,×, in that order with probability  16⁴56⁶   × is ‘not 6’

or  6, ×,×,×,6, 6,×, ×,6,× ,   or  6,6,×,×,6,×,×,×,6,×, or  ……….. all with probability  16⁴56⁶

The  4  sixes could appear on the 10 dice in  a total of  ¹⁰C₄ ways, each way having the same probability, giving

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In general, for  X  ∼  B(n, p)

the probability of  r  successes is

Example:A game of chance has probability of winning 0⋅73  and losing 0⋅27.  Find the probability of winning more than 7 games in 10 games.

Solution:The number of successes is a random variable  X  ~   B(10, 0⋅73), assuming independence of trials.

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Example:For X ~ B(30, 0⋅35), find the probability that  7 < X ≤ 12.

Solution:A moment’s thought shows that we need  P(X = 8,  9, 10, 11 or 12)

using tables for n = 30,  p = 0⋅35

4 D.P. as we are using tables

Example:A bag contains a large number of red and white discs, of which 85% are red. 20 discs are taken from the bag; find the probability that the number of red discs lies between 12 and 17 inclusive.

Solution:As there is a large number of discs in the bag, we can assume that the probability of a red disc remains the same for each trial, p = 0⋅85.

At first glance this looks simple until we realise that the tables stop at probabilities of  0⋅5.

it is worth writing out the numbers

since  X + Y = 20

from tables

Example:A coin is spun 100 times. Find the expected mean and variance of the number of heads.

Solution:X  ∼  B(100, 12)

So we would expect that the probability that the number of heads lies between 40 and 60 inclusive (within 2 standard deviations of the mean) is about 0⋅95.

Example:It is believed that 35% of people like fish and chips. A survey is conducted to verify this. Find the minimum number of people who should be surveyed if the expected number of people who like fish and chips is to exceed 60.

Solution:If  X  is the number of people who like fish and chips in a sample of size n,

Example:Cars pass a particular point at a rate of 5 cars per minute.

(a)  Find the probability that exactly 4 cars pass the point in a minute.

(b)  Find the probability that between at least 3 but fewer than 8 cars pass in a particular minute.

(c)  Find the probability that more than 8 cars pass in 2 minutes.

(d)  Find the probability that more than 3 cars pass in each of two separate minutes.

Solution:Let  X  be the number of cars passing in a minute, then X  ∼  Po(5)

(a)   X  ∼  Po(5)

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or

using tables

(b)   P(at least 3 but fewer than 8) = P(3 ≤ X < 8)

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(c)  We know that the Poisson distribution is uniform, so if a mean of 5 cars pass each minute,

it means that a mean of 10 cars pass in a 2 minute period.

Thus, if  Y  is the number of cars passing in two minutes

using tables

(d)  For probability of more than 3 in one 1 minute period, we have  X  ∼  Po(5)

from tables

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Notice the difference in the parts  (c)  and  (d).   Make sure that you read every question carefully‼

Example:In producing rolls of cloth there are on average 4 flaws in every 10 metres of cloth.

(a)  Find the mean number of flaws in a 30 metre length.

(b)  Find the probability of fewer than 3 flaws in a 6 metre length.

(c)  Find the variance of the number of flaws in a 15 metre length.

Solution:Assuming a Poisson distribution – flaws in the cloth occur singly, independently,

uniformly and randomly.

(a)  If the mean number of flaws in 10 metres is 4,

then the mean number of flaws in 30 metre lengths is 3 × 4 = 12.

(b)  If there are 4 flaws on average in a 10 metre length there will be  610 × 4 = 2⋅4 flaws on

average in a 6 metre length.

If  X  is the number of flaws in a 6 metre length then  X ∼  P_o(2.4).

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(c)  If the mean number of flaws in 10 metre lengths is 4, then the mean number of flaws in 15

metre lengths will be

Since, in a Poisson distribution, the variance equals the mean

the variance in 15 metre lengths is 6.

Example:If the probability of hitting the bull in a game of darts is  120, find the probability of hitting

at least 3 bulls in 50 throws using

(a)  the Binomial distribution

(b)  the Poisson approximation to the Binomial.

Solution:P(at least three bulls) =  1 – P(0, 1 or 2)  =   1 – P( ≤ 2).

(a)  For X ∼ B(50, 0⋅05) the cumulative binomial tables give

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(b)  X ∼  B(50, 0⋅05)  the expected mean    λ = np

We use the approximation  Y ∼ P_o(2.5).

The cumulative Poisson tables for  λ = 2.5  give

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Not surprisingly the answers to parts (a) and (b) are different but not very different.

Example:On average I make 7 typing errors on a page (and that is on a good day!).

(a)  Find the probability that I make more than 10 mistakes on a page.

(b)  In typing 5 pages find the probability that I make more than 10 mistakes on exactly 3 pages.

Solution:

(a)  Assuming single, uniform and independent/random we can use the Poisson distribution

Po(7) and from the cumulative Poisson tables, taking  X  as the number of typing errors

(b)  From (a) we know that the probability of one page with more than 10 errors is 0.0985 and

we take  Y  as the number of pages with more than 10 errors. Thus for 5 pages

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Example: X is a random variable with probability density function

(a)  Find the value of k.

(b)  Find the probability that  12 < x ≤ 1.

(c)  Sketch the probability density function.

Solution:

(a)  The total area between 0 and 2 must be 1

(b)  The probability that  12 < x ≤ 1  is the area between  12  and 1

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(c)  Note that f (x) is zero outside the interval [0, 2]  and this

must be shown on your sketch to gain full marks in the exam.

Example: The random variable X has probability density function

Find the cumulative probability that  X ≤ 3,  i.e. find  F(3).

Solution:  We want  F(3) = P(x ≤ 3)

Notice that we could have drawn a sketch

and found the area of the triangle

Example:  A random variable X  has a probability density function

(a)  Find  F(x).

(b)  Sketch the graph of  F(x).

Solution:

(a)  F(x) = ∫ f(x) dx

0 ≤ x < 1

1 ≤ x < 3

Finding  c′ is more complicated here.

The interval  [0, 1) ends with  F(1) = 110× 1³ =  110,

and so the interval [1, 3) must start with  F(1) =  110.

3 ≤ x < 5

(b)

Example:  A dart is thrown at a dartboard of radius 25 cm. Let X  be the distance from the centre to

the point where the dart lands.

Assuming that the dart is equally likely to hit any point of the board find

(a)  the cumulative probability density function for X.

(b)  the probability density function for X.

Solution:

(a)  F(x) = P(X < x) = P(the dart lands a distance of less than x from the centre)

(b)  f(x) = d(F(x))dx =  ddx ( x²625 ) =  2x625

Example:  Find the mode for a random variable with probability density function

Solution:

We now look at the whole graph in the interval

Example:  Find the median for a random variable with probability density function

Solution:  The median M is given by

Example:Find the probability of more than 20 sixes in 90 rolls of a fair die.

Solution:The exact distribution is binomial  X ∼ B90,  16, where  X  is the number of sixes; but

finding the exact probability would involve much tedious arithmetic.

Note that  n  is large,  np  = 15 > 10   so we can use the normal  N(np, npq)  as an approximation.

using tables

Example:Cars arrive at a motorway filling station at a rate of 18 every quarter of an hour. Find the

probability that at least 23 cars arrive in a quarter of an hour period.

Solution:The exact distribution is Poisson

X  ∼  P_O(18), where  X  is the number of cars arriving in a  14  hour period;

but finding the exact probability would involve much tedious arithmetic.

Note that  n  is large,  λ  = 18 > 10   so we can use the normal  Y ∼ N(18, 18), where  Y  is the

number of cars arriving in a ¼ hour period, as an approximation.

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Example:A large bag contains £1 and £2 coins in the ratio  3 : 1.

A random sample of three coins is taken and their values X₁, X₂ and X₃ are recorded.

Find the sampling distribution for the mean.

Solution:We must first find each sample, its mean and probability

and so the sampling distribution (or probability table) of the mean is

Example:A disease is present in a 23% of a population. A random sample of 30 people is taken and

the number with the disease,  D,  is recorded. What is the sampling distribution of  D?

Solution:The possible outcomes  (values of D)  are

which we recognise as the Binomial distribution,

so the sampling distribution of   D   is    D ~ B(30, 0.23)

Example:A tetrahedral die (one with four faces! – each equally likely) is rolled 40 times and 6 ‘ones’ are observed. Is there any evidence at the 10% level that the probability of a score of  l is less than a quarter?

Notice that the expected mean is 10  (= 40 × 14), and we are really asking if the observed result

(test statistic)  6 is ‘surprisingly low’.

Solution:

H₀:p = 0.25

H₁:p < 0.25.

From H₁ we see that a one-tail test is required, at 10% significance level.

If X is number of ‘ones’ then assuming binomial  X ~ B(40, 0⋅25), from H₀, and using the

cumulative binomial tables

The test statistic (observed value) is  X = 6

P(X ≤ 6 ‘ones’ in 40 rolls) = 0⋅0962 = 9⋅62%from tables

Since 9⋅62% < 10%  the test statistic (observed result) lies in the critical region.

We reject H₀ and conclude that

there is evidence to show that the probability of a score of 1 is lower than

14.

Example:The probability that a footballer scores from a penalty is 0⋅8. In twenty penalties he scores only 13 times. Is there any evidence at the 5% level that the footballer is losing his form?

Solution:

H₀:p = 0⋅8

H₁:p < 0⋅8

If  X  is the number of scores from 20 penalties, then  X ∼ B (20, 0⋅8) – from H₀

The cumulative binomial tables do not deal with p > 0.5 so we must ‘turn the problem round’ and

consider  Y, the number of misses in 20 penalties, where  Y ∼ B (20, 0⋅2).

Observed value is  X = 13

We consider the values    X  ≤  13

Using the cumulative binomial tables,  Y ∼ B (20, 0⋅2)

P(X ≤ 13)   =   P (Y ≥ 7)

       =   1 – P (Y ≤ 6)     =   1 – 0⋅9133     =   0⋅0867 = 8⋅67%

8⋅67% > 5%  (significance level)

⇒  the test statistic (observed result) 7 is not significant (does not lie in the critical region),

Do not reject H₀.

Conclude that there is evidence that the player has not lost his form, or that there is evidence that

the probability of scoring from a penalty is not less than 0⋅8.

Example:A tetrahedral die is manufactured with numbers 1, 2, 3 and 4 on its faces. The manufacturer claims that the die is fair.

All dice are tested by rolling 30 times and recording the number of times a ‘four’ is scored.

(a) Using a 5% significance level, find the critical region for a two-tailed test that the probability of a ‘four’ is 14.

Find critical values which give a probability which is closest to 0⋅025.

(b) Find the actual significance level for this test.

(c) Explain how a die could pass the manufacturer’s test when it is in fact biased.

Solution:

(a)

H₀:p = 0⋅25.

H₁:p ≠ 0⋅25　　  the die is not fair if there are too many or too few ‘fours’

From H₁ we can see that a two-tailed test is needed, significance level 2⋅5% at each end.

Let X be the number of ‘4’s in 30 rolls. From H₀ we have a binomial distribution,  X ~ B(30, 0⋅25).

We shall reject the hypothesis if the observed result lies in either half of the critical region each

half having a significance level of 2⋅5%;

For a two tail test, find the values of X which give a probability closest to 2⋅5% at each end.

Using cumulative binomial tables for  X ∼ B (30, 0⋅25):

for the lower critical valuefrom tables

P(X ≤ 2) = 0⋅0106      (2.5%  – 1.06%   =  1.44%)

P(X ≤ 3) = 0⋅0374     (3.74% – 2.5%    =  1.24%)

X ≤ 3 gives the value closest to 2⋅5%, so  X = 3 is lower critical value

and for the higher critical valuefrom tables

P(X ≥ 13)  =   1– P(X ≤ 12) = 1 – 0.9784 = 0.0216     (2.5%  – 2.16%  =  0.34%)

P(X ≥ 12)  =   1– P(X ≤ 11) = 1 – 0.9493 = 5.07%   (5.07%   – 2.5%  =  2.57%)

X ≥ 13 gives the value closest to 2⋅5%, so  X = 13 is higher critical value

Thus the critical region is     X ≤ 3   or   X ≥ 13.

(b)

The actual significance level is  0⋅0374 + 0⋅0216  =   0⋅0590  =   5⋅90%.      to 3 S.F.

(c)

The die could still be biased in favour of, or against, one of the other numbers.

Example:Cars usually arrive at a motorway filling station at a rate of 3 per minute. On a Tuesday morning cars are observed to arrive at the filling station at a rate of 5 per minute.  Is there any evidence at the 10% level that this is an unusually busy morning?

Solution:

H₀:λ = 3

H₁:λ > 3　　 unusually busy would mean that  λ  would increase