STATISTICS

quants_mock.docx

Home >Mathematics homework help >STATISTICS

Quantitative Methods - Mock

Question one (6 Marks)

Three coins are tossed together.

a) List all possible combinations of heads (H) and tails (T).
	(2 marks)

b) What is the probability of having exactly one head?
	(2 marks)

c) What is the probability of at least 2 heads?
	(2 marks)

Question two (8 marks):

In a game of selecting a 3-digit number, you pay 50p to play a game.

(Each selected digit can be any number from 0 to 9 and may be selected more than once.)

If you select the correct number, you win £250

a) How many different selections are possible?
	(2 marks)

b) What is the probability of winning?
	(2 marks)

c) If you win, what is your profit?
	(1 marks)

a) d) Find the expected value of winning and say whether you would play this game.
	(3 marks)

Question three (6 marks)

The table below summarises the job applications made by some graduates during their last year:

Was the application successful?

Yes No

Male graduates 263 148

Female graduates 178 122

If you were to randomly select a graduate:

a) What is the probability that an application was not successful? (2 marks)

b) What is the probability the application was made by a male graduate or was successful? (4 marks)

Question four (20 marks):

X is a random variable which represents the weight in kg of crates of apples in the local market. Assume it has a normal distribution with a mean of 50 and variance of 49.

a) What proportion of the crates weigh over 80kg?
	(4 marks)

b) 68% of the crates will weigh between [?, ?].
	(2 marks)

c) What proportion of the crates will weigh between 40kg and 53kg? Interpret it.
	(6 marks)

d) A sample of 25 crates was randomly drawn. What is the probability that the sample mean is between 47kg and 54kg?
	(8 marks)

Question five (15 marks):

For a sample of 201 households from Norwich it was found that the average monthly expenditure on food was £500 with sample variance £2,500.

a) Using the given formula construct a 95% confidence interval for the average monthly expenditure on food. (4 marks)

b) You were informed that the monthly expenditure on food is £510; you believe that this is too high. Conduct a test in order to prove that you are right.

(6 marks)

c) You think that household monthly expenditure on food would be £505, test your guess. (5 marks)

Question six (20 marks):

The ‘golden ratio’ 1 to 0.618. It is claimed that rectangles whose sides are in this ratio are pleasing to the eye (i.e. they look nice).

An experiment was conducted to see if this claim is true of the shape of paintings. That is, the ratio of the length to the width of a picture frame. Twenty people were asked to look at different sized picture frames and choose the most pleasing shape. The data below are the width ratios they chose as part of the study:

0.632	0.690	0.606	0.570	0.749	0.670	0.672	0.628
0.844	0.654	0.615	0.606	0.601	0.576	0.614	0.553
0.668	0.693	0.609	0.901

The significance level is 10%.

a) State formally and clearly the hypotheses. (2 marks)

b) Is this a one or two-tail test? Why? (2 marks)

c) Calculate the t-statistic (6 marks)

d) State the critical value (4 marks)

e) Is the claim true? Briefly explain your answer. (6 marks)

Statistical Formulae

Variance:

s 2 = Σx2 - (Σx)2/n

n – 1

Z score:

Z = X – mean

Coefficient of variation:

CV = S x 100

Mean

Probability (equally likely):

P(Event) = Number of outcomes in that the Event occurs

Total number of possible outcomes

Probability (not equally likely):

P(Event) = Number of times the Event of interest occurs

Total number of times

Complementary events:

P(A) + P(Not A) = 1

P(Not A) = 1 – P(A)

Combining events:

P(A or B) = P(A) + P(B) – P(A AND B)

P(A AND B) = P(A) + P(B) – P(A OR B)

Conditional probability:

P(A|B) = P(A AND B) or P(B|A) = P(A AND B)

P(B) P(A)

Can be written as …

P(A AND B) = P(A|B) P(B) or P(A AND B) = P(B|A) P(A)

An independent event:

P(A AND B) = P(A) P(B)

Expected value:

μ = E(x) = ΣxP(x)

Expected value of x2:

E(x2) = Σx2P(x)

Binomial combinations:

nCx = n ! .

x!(n – x)!

Binomial probability:

P(x) = nCx p x(1 – p) n-x

Where:

nCx = n ! .

x !(n – x)!

Expected value and variance of a continuous distribution:

The formula for mean and the variance of a uniformly distributed random variable:

f(x) = 1 .

b – a

a < x < b

The mean is:

a + b

μ = 2

The variance is:

σ2 = (b – a)2

Standardising a normal random variable:

Z = X – μ

The distribution of sample mean:

95% confidence interval:

To test H0: =0:

Calculate pooled variance of two samples.

To test H0: 1=2:

)

(

)

(

)

(

)

(

025