S1 Statistics 1

Example:When a die is rolled, we say that the probability of each number is 16. This is a statistical model, but the assumption that each face is equally likely might not be true. Suppose the die is weighted to increase the chance of a six. We might then find, after experimenting, that the probability of a six is 14 and the probability of a one is 112, with the probability of other faces remaining at 16, all adding up to 1. In this case we have refined, or improved, the model to give a truer picture.

Example:The heights of a large group of adults are measured. The mean is 172⋅3 cm and the standard deviation is 12⋅4 cm.

It could be thought that the general shape of the histogram can be modelled by the curve

This might not give a true picture, in which case we would have to change the equation, or refine the model.

Example:The number of M&Ms is counted in several bags, and recorded in the frequency distribution/table below:

Example:0 ≤ x < 5, 5 ≤ x < 9, 9 ≤ x < 16, etc.

Note that each interval starts at the point where the previous interval ended, but 5, for example, goes in the second interval 5 ≤ x < 9.

Example:100 people were asked how many DVDs they owned. The results are shown below.

(a) What are the class boundaries for the interval 12 to 14?

(b) What is the area of the bar for the class 12 to 14? Compare this with the frequency for this class.

(c) Fill in the blanks in the frequency table. Check that your frequencies add up to 100.

(d) Find the number of people who own between 15 and 24 DVDs, inclusive.

Solution:

(a) Class boundaries for the interval 12 to 14 are 11⋅5 to 14⋅5

(b) Class width of bar for 12 to 14 is 14⋅5 – 11⋅5 = 3, height of bar = 2⋅5, ⇒ area = 3 × 2⋅5 = 7⋅5 which is half the frequency, 15.

Thus the frequency is always twice the area.

(c) 15 to 16 is 14⋅5 to 16⋅5 width 2, height 4⋅5 ⇒ area 9 ⇒ frequency = 2× 9 = 18

17 is 16⋅5 to 17⋅5 width 1, height 8⋅5 ⇒ area 8⋅5 ⇒ frequency = 2× 8⋅5 = 17

18 to 21 is 17⋅5 to 21⋅5 width 4, height 4 ⇒ area 16 ⇒ frequency = 2 × 16 = 32

22 to 30 is 21⋅5 to 30⋅5 width 9, height 1 ⇒ area 9 ⇒ frequency = 2× 9 = 18

Check 15 + 18 + 17 + 32 + 18 = 100, which is correct

(d) 15 to 24 is 14⋅5 to 24⋅5

14⋅5 to 16⋅5 and 16⋅5 to 17⋅5 and 17⋅5 to 21⋅5 gives f = 18 + 17 + 32

21⋅5 to 24⋅5 has width 3 and height 1 so area = 3 ⇒ f = 2 × 3 = 6

⇒ number between 15 and 24 inclusive is 18 + 17 + 32 + 6 = 73.

Example:A histogram is to be drawn from the following data:

The width of the 2 − 4 interval is 6 cm, and its height is 8 cm.

Find the widths and heights, in cm, for the 5 − 6, the 7 − 12, and the 13 − 16 bars.

Solution:

(a) The 2 − 4 interval has width 6 cm, height 8 cm ⇒ area = 6 × 8 = 48 cm 2.

The 2 − 4 interval is 1⋅5 to 4⋅5, so has width 3 ∝ 6 cm and frequency 15 ∝ 48 cm2.

(b) The 5 − 6 interval is 4⋅5 − 6⋅5, so has width 2, which is 23 of the width of the 2 to 4 interval, so its width (in cm) is 23× 6 = 4 cm.

The 5 − 6 bar has a frequency of 6, which is 615 of the frequency of the 2 − 4 bar, so the area of the 5 − 6 bar is 615× 48 = 19⋅2 cm2,

⇒ height is area ÷ width = 19⋅2 cm2 ÷ 4 cm = 4⋅8 cm.

(c) We can now fill in the table below, working from left to right.

Example:A grouped frequency table for the weights of adults has the following entries:

In a histogram, the bar for the class 50 – 60 kg is 2 cm wide and 9 cm high.

Find the width and height of the bar for the 70 – 85 kg class.

Solution:50 – 60 is usually taken to mean 50 ≤ weight < 60

The width of the 50 – 60 class is 10 kg ≡ 2 cm

The area of the 50 – 60 bar is 2 cm × 9 cm = 18 cm2 ⇔ frequency is 60

Answer the width of 70 – 85 kg bar is 3 cm, and the height is 2 cm.

Example:Find the mean for the following table showing the number of children per family.

Solution:

Σ x_if_i = 142, and N = Σ f_i = 56

⇒  x̅ = 14256 = 2⋅54 to 3 S.F.

Example:The table shows the numbers of children in prep school classes in a town.

Solution:

Σ x_i f_i = 1197, and N = Σ f_i = 54

⇒  x̅ = 119754 = 22⋅2 to 3 S.F.

Example:Use the coding t_i = x_i−16510 to find the mean weight for the following distribution.

⇒  t̅ = 11106

and  t_i = x_i−16510  ⇒  x̅ = 10 t̅ + 165 = 10 × 11106 + 165 = 166⋅0377358

⇒  mean weight is 166⋅04 kg to 2 D.P.

Example:For the 90th percentile, find

Example:The ages of 11 children are given below.

Q₁ = 12, Q₂ = 14 and Q₃ = 17.

Outliers are values outside the range Q₁ – 1⋅5 × (Q₃ – Q₁) to Q₃ + 1⋅5 × (Q₃ – Q₁).

Find any outliers, and draw a box plot.

Solution:Lower boundary for outliers is 12 – 1⋅5 × (17 – 12) = 4⋅5

Upper boundary for outliers is 17 + 1⋅5 × (17 – 12) = 24⋅5

⇒ 3 and 26 are the only outliers.

To draw a box plot, put crosses at 3 and 26 (outlilers), and draw the lines to 6 (the lowest value which is not an outlier), and to 21 (the highest value which is not an outlier).

Example:Find the mean and standard deviation for the following distribution.

Here the x-values are nasty, but if we change them to form t_i = x_i−2105 then the arithmetic in the last two columns becomes much easier.

the mean of t is t̅ = 1N∑t_if_i = 3117 = 139

and the variance of t is (sd)_t² = 1N∑t_i²f_i − t̅² = 141117 − (139)² = 1⋅204470743

(sd)_t = √1∙204470743 = 1∙097483824 = 1.10 to 3 S.F.

To find x̅, using t_i = x_i−2105 , ⇒ x̅ = 5t̅ + 210 = 5 × 139 + 210 = 210.13 to 2 D.P.

To find the standard deviation of x

(sd)_x = 5(sd)_t = 5 × 1⋅0974838… = 5⋅49 to 3 S.F.

We would need to multiply the variance by 5² = 25

⇒  (sd)_x² = 25(sd)_t² = 25 × 1⋅204470743 = 30⋅1 to 3 S.F.

Example:One class of 20 boys has a mean mass of 62 kg with a standard deviation of 3 kg.

A second class of 30 boys has a mean mass of 60 kg with a standard deviation of 4 kg.

Find the mean and standard deviation of the combined class of 50 boys.

Example:In the launch of a rocket, the probability of an electrical fault is 0⋅2. If there is an electrical fault the probability that the rocket crashes is 0⋅4, and if there is no electrical fault the probability that the rocket crashes is 0⋅3.

Draw a tree diagram. The rocket takes off, and is seen to crash. What is the probability that there was an electrical fault?

Solution:Let E be an electrical fault, and C be a crash.

We want to find P(E | C).

P(E | C) = P(E∩C)P(C)

P(E∩C) = 0⋅2 × 0⋅4 = 0⋅08

and  P(C) = 0⋅2 × 0⋅4 + 0⋅8 × 0⋅3 = 0⋅32

⇒  P(E | C) = 0⋅080⋅32 = 0⋅25

Example:A red die and a green die are rolled and the total score recorded.

A is the event ‘total score is 7’,  B is the event ‘green score is 6’ and C is the event ‘total score is 10’.

Show that  A and B are independent, but  B and C are not independent.

Solution:The events A, B and C are shown on the diagram.

P(A) = 636 = 16, P(B) = 636 = 16

and  P(A ∩ B) = 136 from diagram

P(A) × P(B) = 16 × 16 = 136 = P(A ∩ B)

⇒ A and B are independent.

P(B) = 636 = 16, P(C) = 336 = 112

and  P(B ∩ C) = 136 from diagram

P(B) × P(C) = 16 × 112 = 172 ≠ P(B ∩ C)

⇒ B and C are not independent

Example:A and B are independent events.  P(A) = 0⋅5 and P(A ∩ B′) = 0⋅3. Find P(B).

Solution:  P(A) = 0⋅5 and P(A ∩ B′) = 0⋅3

⇒ P(A ∩ B) = P(A) − P(A ∩ B′) = 0⋅5 – 0⋅3

⇒ P(A ∩ B) = 0⋅2

But  P(A ∩ B) = P(A) × P(B) since A and B are independent

⇒ P(A ∩ B) = 0⋅5 × P(B)

⇒ 0⋅5 × P(B) = 0⋅2

⇒ P(B) = 0∙20∙5 = 0⋅4

Example:P(A) = 0⋅3, P(B) = 0⋅6 and P(A′∩ B′) = 0⋅1.

Prove that A and B are mutually exclusive.

Solution:P(A′∩ B′) is shaded in the diagram

⇒ P(A′∩ B′) = 1 – P(A ∪ B)

⇒ P(A ∪ B) = 1 – 0⋅1 = 0⋅9

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ 0.9 = 0⋅3 + 0⋅6 – P(A ∩ B)

⇒ P(A ∩ B) = 0

⇒ A and B are mutually exclusive.

Example:A bag contains  5 Red beads, 7 Yellow beads, and 6 White beads.  Three beads are drawn without replacement from the bag. Find the probability that there are 2 Red beads and 1 Yellow bead.

Solution:  These beads can be drawn in any order,   RRY,  RYR,  YRR

⇒ P(RRY or RYR or YRR)

= P(RRY) + P(RYR) + P(YRR)

= 518 × 417 × 716 + 518 × 717 × 416 + 718 × 517 × 416 = 35408.

You must always remember the possibility of more than one order. In rolling four DICE, exactly TWO SIXES can occur in six ways:

SSNN,  SNSN,  SNNS,  NSSN,  NSNS,  NNSS, each of which would have the same probability 16² × 56²

= 251296

and so the probability of exactly two sixes with four dice is  6 × 251296 = 25216.

Example:Investigate the effect of the coding  t = x−ka  on the PMCC.

Solution:r_xy = ∑(x_i−x̅)(y−y̅)√∑(x_i−x̅)² ∑(y_i−y̅)²

t_i = x_i−ka ⇒ x_i = at_i + k and x̅ = at̅ + k

⇒  r_xy = ∑(at_i+ k−(at̅ + k))(y−y̅)√∑(at_i+ k−(at̅ + k))² ∑(y_i−y̅)²

 =  ∑a(t_i−t̅)(y−y̅)√∑a²(t_i−t̅)² ∑(y_i−y̅)²

 =  a∑(t_i−t̅)(y−y̅)√a²√∑(t_i−t̅)² ∑(y_i−y̅)²

 =  ∑(t_i−t̅)(y−y̅)√∑(t_i−t̅)² ∑(y_i−y̅)²

 =  r_ty

Example:Bleep tests are used to measure people’s fitness. A higher score means a higher level of fitness. The heart rate, p beats per minute, and bleep score, s, for 12 people were recorded and coded, using  x =  p – 60  and  y = 10s – 50.

Σ x = 171, Σ y = 270, Σ x² = 4477, Σ y² = 13540, Σ xy = 1020.

(a) Find the PMCC between x and y.

(b) Write down the PMCC between p and s.

(c) Explain why your answer to (b) might suggest that there is a linear relationship between p and s.

(d) Interpret the significance of the PMCC.

Solution:(a) S_xy = ∑x_iy_i − 1N(∑x_i)(∑y_i) = 1020 – 171×27012 = –2827⋅5

S_xx = ∑x_i² − 1N(∑x_i)² = 4477 – 171²12 = 2040⋅25

S_yy = ∑y² − 1N(∑y_i)² = 13540 – 270²12 = 7465

⇒  r = S_xy√S_xxS_yy = −2827∙5√2040∙25×7465 = – 0⋅7245127195 = – 0⋅725 to 3 S.F.

(b) As coding has no effect on the PMCC, the product moment correlation coefficient for p and s is also  – 0⋅725,  to 3 S.F.

(c) r = – 0⋅725  is ‘quite close’ to –1, and therefore the points on a scatter diagram would lie close to a straight line

⇒ there is evidence of a linear relation between p and s.

(d) Always give both answers − the technical describing the correlation, and the wordy one which copies out the question. Then, whatever the examiner wants, you should get the mark(s).

There is negative correlation between p and s, which means that

as heart rate increases, the bleep score decreases, or

people with higher heart rate tend to have lower bleep scores.

Example: A local authority is investigating the cost of reconditioning its incinerators. Data from 10 randomly chosen incinerators were collected. The variables monitored were the operating time x (in thousands of hours) since last reconditioning and the reconditioning cost y (in £1000). None of the incinerators had been used for more than 3000 hours since last reconditioning.

The data are summarised below,

(a) Find the equation of the regression line of y on x.

(b) Give interpretations of a and b.

Solution:

(a) x̅ = 25∙010 = 2⋅50,  y̅ = 50∙010 = 5⋅00,

S_xy = 130⋅64 – 25∙0×50∙010 = 5⋅64,  S_xx = 65.68 – (25∙0)²10 = 3⋅18

⇒ b = S_xyS_xx = 5.643.18 = 1⋅773584906

⇒ a = y̅ – bx̅ = 5⋅00 – 1⋅773584906 × 2⋅50 = 0⋅5660377358

⇒ regression line equation is  y = 0⋅566 + 1⋅77x  to 3 S.F.

(b) a is the cost in £1000 of reconditioning an incinerator which has not been used, so the cost of reconditioning an incinerator which has not been used is £566. (a is the value of y when x is zero)

b is the increase in cost (in £1000) of reconditioning for every extra 1000 hours of use, so it costs an extra £1774 to recondition an incinerator for every 1000 hours of use. (b is the gradient of the line)

Example: A drilling machine can run at various speeds, but in general the higher the speed the sooner the drill needs to be replaced.

Over several months, 15 pairs of observations of speed, s revolutions per minute, and the life of the drill, h hours, are collected. For convenience the data are coded so that s = x + 20 and h = 5y + 100

The regression line of y on x is calculated as y = 57⋅14 − 3⋅090x.

Find the regression line of h on s.

Solution: First find x and y in terms of s and h.

s = x + 20 and h = 5y + 100

⇒ x = s − 20 and y=h−1005.

Then substitute into the equation of the regression line, y = 57⋅14 − 3⋅090x .

⇒ h−1005 = 57⋅14 − 3⋅090(s − 20)

⇒ h − 100 = 285⋅7 − 15⋅45s + 309⋅0

⇒ h = 694⋅7 − 15⋅45s is the regression line of h on s.

Example: The weights, w, and heights, h of four people are shown in the table below. (In practice there would be more people, but it is easier to show the method with a small number of pairs.)

(a) Find the mean weight, mean height and the PMCC.

(b) A fifth person joins the group with weight 77 kg and height 171 cm. Find the new value of S_hh.

(c) Given that the addition of the fifth person increases the value of S_wh, what is the effect of the fifth person on the gradient of the regression line of w on h?

Solution:

(a)

w=2964 = 74 and h=6844 = 171

S_wh= 50658 − 296×6844 = 42, S_ww= 21970 − 296²4 = 66, S_hh= 116994 − 684²4 = 30

⇒ r = 42√66×√30 = 0⋅943879807 = 0⋅ 944 to 3 S.F.

(b) To answer part (b) we use the formulae in the second column.

S_hh = 4∑1(h_i − h)²

= (173−171)² + (167−171)² + (174−171)² + (170−171)²

The height, h₅ = 171 cm, of the fifth person does not change the mean height, h = 171 cm. To find the new S_hh after the fifth person joins the group, we have to add the term

(h₅ −h)² = (171 − 171)² = 0.

and so zero is added to S_hh, and the new S_hh = the old S_hh .

(c) The regression line of w on h is w = a + bh.

The gradient is b = S_whS_hh .

We are told that the addition of the fifth person increases the value of S_wh, and we have shown that S_hh is not changed

⇒ the new gradient = biggersame , and is therefore bigger than the original gradient for four people.

Example: A class of 25 students takes a physics exam and a mathematics exam. The equation of the regression line of physics marks, p, against mathematics marks, m, is found using

S_pp = 1241, S_pm = 649 and S_mm = 847,  p= 57 and m= 65.

b= S_pmS_mm = 649847 = 0 ∙766

(a) Which is the explanatory variable?

(b) A pupil who takes the exams at a later date scores 62 in physics and 65 marks in mathematics.

(i) Find the new value of S_mm.

(ii) What can you say, without detailed calculation, about the new values of S_pm and S_pp?

Solution:

(a) m is the explanatory variable (and p is the response variable).

(b) (i)  The addition of an extra mark, 65, which is equal to the mean, m = 65, of the original group, will leave the mean unchanged.

S_mm = ∑(m_i−m)²

The result, 65 in mathematics, will give an extra term in S_mm.

The extra term = (65 −m)² = (65 −65)² = 0

Adding 0 to S_mm = ∑(m_i−m)² has no effect

⇒  S_mm is unchanged,  ⇒  new S_mm = 847

(ii)  new S_pm = ∑(p_i−p_new)(m_i−m); and new S_pp = ∑(p_i−p_new)²

As we do not know the new value p_new, we cannot, without detailed calculation, say anything about S_pm or S_pp.

Example:

is the probability distribution (table) for the random variable, X, the total score on two dice.

Example: A fair die is rolled and the score recorded.

(a) Find the expected mean and variance for the score, X.

(b) A ‘prize’ is awarded which depends on the score on the die. The value of the prize is $Z = 3X – 6. Find the expected mean and variance of Z.

Solution:

(a)

(b) Z = 3X – 6

Example: A die is rolled until a 6 is thrown, or until the die has been rolled 4 times. Find the expected mean and standard deviation of the number of times the die is rolled.

Solution: Expected mean  = µ = E[X] = ∑x_ip_i

and    expected variance = σ² = E[X²] − (E[X])² = ∑x_i²p_i−µ²

Answer : Expected mean number of rolls is 3⋅11, with standard deviation 1⋅17

Example: The random variable X is defined as the score on a single die. X is a discrete uniform distribution on the set {1, 2, 3, 4, 5, 6}

The probability distribution is

Example: X is the score on a fair 10 sided spinner. Define Y = 5X + 3.

Find the mean and variance of Y.

Y is the distribution {8, 13, 18, … 53}, all with the same probability 110 .

Solution: X is a (standard) discrete uniform distribution on the set {1, 2, 3, …, 10}

Example: P(Z < –1⋅23)