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1. A company has a large number of regular users logging onto its website. On average 4 users every hour fail to connect to the company’s website at their first attempt.
(a) Explain why the Poisson distribution may be a suitable model in this case. (1)
Find the probability that, in a randomly chosen 2 hour period,
(b) (i) all users connect at their first attempt,
(ii) at least 4 users fail to connect at their first attempt. (5)
The company suffered from a virus infecting its computer system. During this infection it was found that the number of users failing to connect at their first attempt, over a 12 hour period, was 60.
(c) Using a suitable approximation, test whether or not the mean number of users per hour who failed to connect at their first attempt had increased. Use a 5% level of significance and state your hypotheses clearly. (9) (Total 15 marks)
2. A café serves breakfast every morning. Customers arrive for breakfast at random at a rate of 1 every 6 minutes.
Find the probability that
(a) fewer than 9 customers arrive for breakfast on a Monday morning between 10 am and 11 am. (3)
The café serves breakfast every day between 8 am and 12 noon.
(b) Using a suitable approximation, estimate the probability that more than 50 customers arrive for breakfast next Tuesday. (6) (Total 9 marks)
3. An administrator makes errors in her typing randomly at a rate of 3 errors every 1000 words.
(a) In a document of 2000 words find the probability that the administrator makes 4 or more errors. (3)
The administrator is given an 8000 word report to type and she is told that the report will only be accepted if there are 20 or fewer errors.
(b) Use a suitable approximation to calculate the probability that the report is accepted. (7) (Total 10 marks)
4. A web server is visited on weekdays, at a rate of 7 visits per minute. In a random one minute on a Saturday the web server is visited 10 times.
(a) (i) Test, at the 10% level of significance, whether or not there is evidence that the rate of visits is greater on a Saturday than on weekdays. State your hypotheses clearly.
(ii) State the minimum number of visits required to obtain a significant result. (7)
(b) State an assumption that has been made about the visits to the server. (1)
In a random two minute period on a Saturday the web server is visited 20 times.
(c) Using a suitable approximation, test at the 10% level of significance, whether or not the rate of visits is greater on a Saturday. (6) (Total 14 marks)
5. (a) State the condition under which the normal distribution may be used as an approximation to the Poisson distribution. (1)
7. A manufacturer produces large quantities of coloured mugs. It is known from previous records that 6% of the production will be green.
A random sample of 10 mugs was taken from the production line.
(a) Define a suitable distribution to model the number of green mugs in this sample. (1)
(b) Find the probability that there were exactly 3 green mugs in the sample. (3)
A random sample of 125 mugs was taken.
(c) Find the probability that there were between 10 and 13 (inclusive) green mugs in this sample, using
(i) a Poisson approximation, (3)
(ii) a Normal approximation. (6) (Total 13 marks)
8. A teacher thinks that 20% of the pupils in a school read the Deano comic regularly.
He chooses 20 pupils at random and finds 9 of them read Deano.
(a) (i) Test, at the 5% level of significance, whether or not there is evidence that the percentage of pupils that read Deano is different from 20%. State your hypotheses clearly.
(ii) State all the possible numbers of pupils that read Deano from a sample of size 20 that will make the test in part (a)(i) significant at the 5% level. (9)
The teacher takes another 4 random samples of size 20 and they contain 1, 3, 1 and 4 pupils that read Deano.
(b) By combining all 5 samples and using a suitable approximation test, at the 5% level of significance, whether or not this provides evidence that the percentage of pupils in the school that read Deano is different from 20%. (8)
(c) Comment on your results for the tests in part (a) and part (b). (2) (Total 19 marks)
9. The random variable X is the number of misprints per page in the first draft of a novel.
(a) State two conditions under which a Poisson distribution is a suitable model for X. (2)
The number of misprints per page has a Poisson distribution with mean 2.5. Find the probability that
(b) a randomly chosen page has no misprints, (2)
(c) the total number of misprints on 2 randomly chosen pages is more than 7. (3)
The first chapter contains 20 pages.
(d) Using a suitable approximation find, to 2 decimal places, the probability that the chapter will contain less than 40 misprints. (7) (Total 14 marks)
10. The random variables R , S and T are distributed as follows
R ~ B(15, 0.3), S ~ Po(7.5), T ~ N(8, 2 2 ).
Find
(a) P( R = 5), (2)
(b) P( S = 5), (1)
The carpet fitter orders a total of 355 m^2 of the carpet for the whole hotel.
(d) Using a suitable approximation, find the probability that this total area of carpet contains 22 or more defects. (6) (Total 14 marks)
13. The random variable R has the binomial distribution B(12, 0.35).
(a) Find P( R ≥ 4). (2)
The random variable S has the Poisson distribution with mean 2.71.
(b) Find P( S ≤ 1). (3)
The random variable T has the normal distribution N(25, 5 2 ).
(c) Find P( T ≤ 18). (2) (Total 7 marks)
14. (a) Write down the condition needed to approximate a Poisson distribution by a Normal distribution. (1)
The random variable Y ~ Po(30).
(b) Estimate P( Y > 28). (6) (Total 7 marks)
The random variable T has the normal distribution N(25, 5 2 ).
(c) Find P( T ≤ 18). (2) (Total 7 marks)
15. A farmer noticed that some of the eggs laid by his hens had double yolks. He estimated the probability of this happening to be 0.05. Eggs are packed in boxes of 12.
Find the probability that in a box, the number of eggs with double yolks will be
(a) exactly one, (3)
(b) more than three. (2)
A customer bought three boxes.
(c) Find the probability that only 2 of the boxes contained exactly 1 egg with a double yolk. (3)
The farmer delivered 10 boxes to a local shop.
(d) Using a suitable approximation, find the probability that the delivery contained at least 9 eggs with double yolks. (4)
The weight of an individual egg can be modelled by a normal distribution with mean 65 g and standard deviation 2.4 g.
(e) Find the probability that a randomly chosen egg weighs more than 68 g. (3) (Total 15 marks)
16. An athletics teacher has kept careful records over the past 20 years of results from school sports days. There are always 10 competitors in the javelin competition. Each competitor is allowed 3 attempts and the teacher has a record of the distances thrown by each competitor at each attempt. The random variable D represents the greatest distance thrown by each competitor and the random variable A represents the number of the attempt in which the competitor achieved their greatest distance.
(a) State which of the two random variables D or A is continuous. (1)
A new athletics coach wishes to take a random sample of the records of 36 javelin competitors.
1. (a) Connecting occurs at random/independently, singly or at a constant rate B1 1
Note B1 Any one of randomly/independently/singly/constant rate. Must have context of connection/logging on/fail
(b) Po (8) B Note B1 Writing or using Po(8) in (i) or (ii)
(i) P( X = 0) = 0.0003 M1 A Note M1 for writing or finding P( X = 0) A1 awrt 0.
(ii) P( X ≥ 4) = 1 – P( X ≤ 3) M = 1 – 0. = 0.9576 A1 5 Note M1 for writing or finding 1 – P( X ≤ 3) A1 awrt 0.
(c) H0 : λ = 4 (48) H1 : λ > 4 (48) B
N(48, 48) M1 A
Method 1 Method 2
P( X > 59.5) =
P
Z
x − 0. 5 − 48
= 1.6449 M1 M1 A
= P( Z > 1.66)
= 0.0485 x = 59.9 A 0.0485 < 0. Reject H 0. Significant. 60 lies in the Critical region M The number of failed connections at the first attempt has increased. A1 ft 9
Note B1 both hypotheses correct. Must use λ or μ M1 identifying normal A1 using or seeing mean and variance of 48 These first two marks may be given if the following are seen in the standardisation formula : 48 and 48 or awrt 6. M1 for attempting a continuity correction (Method 1: 60 ± 0.5 / Method 2: x ± 0.5) M1 for standardising using their mean and their standard deviation and using either Method 1 [59.5, 60 or 60.5. accept ± z .] Method 2 [( x ± 0.5) and equal to a ± z value)
A1 correct z value awrt ±1.66 or ± 48
, or 1. 6449 48
x
A1 awrt 3 sig fig in range 0.0484 – 0.0485, awrt 59. M1 for “reject H 0 ” or “significant” maybe implied by “correct contextual comment” If one tail hypotheses given follow through “their prob” and 0.05, p < 0. If two tail hypotheses given follow through “their prob” with 0.025, p < 0. If one tail hypotheses given follow through “their prob” and 0.95, p > 0. If two tail hypotheses given follow through “their prob” with 0.975, p > 0. If no H 1 given they get M A1 ft correct contextual statement followed through from their prob and H 1. need the words number of failed connections/log ons has increased o.e. Allow “there are more failed connections” NB A correct contextual statement alone followed through from their prob and H 1 gets M1 A [15]
2. (a) X ~ Po(10) B
P( X < 9) =P( X ≤ 8) M
= 0.3328 A1 3
Note B1 for using Po(10)
M1 for attempting to find P( X ≤ 8): useful values
P( X ≤ 9) is 0.4579(M0), using Po(6) gives 0.8472, (M1).
A1 awrt 0.333 but do not accept
3. (a) X = the number of errors in 2000 words so X ~Po(6) B P( X ≥ 4) = 1 – P( X ≤ 3) M = 1 – 0.1512 = 0.8488 awrt 0.849 A1 3 Note B1 for seeing or using Po(6) M1 for 1 – P( X ≤ 3) or 1 – [P( X = 0) + P( X = 1) + P( X = 2) + P( X = 3)] A1 awrt 0. SC If B(2000, 0.003) is used and leads to awrt 0.849 allow B0 M1 A If no distribution indicated awrt 0. scores B1M1A1 but any other awrt 0. scores B0M1A
(b) Y = the number of errors in 8000 words. Y ~ Po(24) so use a Normal approx M
Y ≈~ N(24,
2 24 ) A
Require P(Y ≤ 20) = P (^)
Z M
M
= P( Z < – 0.714…) A
= 1 – 0.7611 M
= 0.2389 awrt (0.237~0.239) A1 7 [N.B. Exact Po gives 0.242 and no ± 0.5 gives 0.207]
Note
1 st^ M1 for identifying the normal approximation
1 st^ A1 for [mean = 24] and [sd = 24 or var = 24] These first two marks may be given if the following are seen in the standardisation formula :
or awrt 4.
2 nd^ M1 for attempting a continuity correction (20/ 28 ± 0.5 is acceptable)
3 rd^ M1 for standardising using their mean and their standard deviation.
2 nd^ A1 correct z value awrt ± 0.71 or this may
be awarded if see 24
or 24
4 th^ M1 for 1 – a probability from tables (must have an answer of < 0.5)
3 rd^ A1 answer awrt 3 sig fig in range 0.237 – 0. [10]
4. (a) (i) H 0 : λ = 7 H 1 : λ > 7 B
X = number of visits. X ~ Po(7) B P ( X ≥ 10) = 1 – P( X ≤ 9) 1 – P( X ≤ 10) = 0.0985 M = 0.1695 1 – P( X ≤ 9) = 0. CR X ≥ 11 A 0.1695 > 0.10, CR X ≥ 11 Not significant or it is not in the critical region or do not reject H 0 M The rate of visits on a Saturday is not greater/ is unchanged A1 no ft
(ii) X = 11 B1 7
(b) (The visits occur) randomly/ independently or singly or constant rate B1 7
(b) P( X = 5) = P( X ≤ 5) − P( X ≤ 4) or 5!
7 5 e−^7 M
= 0.1277 A1 2
awrt 0.
(c) P ( X > 181 ) ≈ P ( Y ≥ 181.5 ) where Y ~N ( 168, 168) N ( 168, 168) B
P z
±0.5 M
stand with μ and σ M
Give A1 for 1.04 or correct expression A = P (z ≥ 1.04)
=1 − 0.8508 M attempt correct area
1 − p where p > 0.
= 0.1492 A1 6
awrt 0. [11]
7. (a) Binomial B1 1
Let X represent the number of green mugs in a sample
(b) X ~ B ( 10, 0.06) B may be implied or seen in part a
P ( X = 3) = 10 C 3 (0.06)^3 (0.94)^7 M (^10) C 3 ( p )^
(^3) (l – p ) 7
0.016808.... A1 3
awrt 0.
(c) Let X represent number of green mugs in a sample of size 125 (i) X ~ P 0 (125 × 0.06 = 7.5) may be implied B
(10 ≤ X ≤ 13) = P( X ≤ 13) – P(X ≤ 9) M 0.9784 – 0. = 0.2020 A1 3 awrt 0.
(ii) P(10 ≤ X ≤ 13) ≈ P(9.5 ≤ Y ≤ 13.5) where Y ∪ N(7.5, 7.05) 7.05 B 9.5, 13.5 B
P z
±0.5 M
stand. M both values or both correct expressions.
= P(0.75 ≤ z ≤ 2.26) A awrt 0.75 and 2.
= 0.2147 A1 6 awrt 0.214 or 0. [13]
8. (a) (i) Two tail B1 B
H 0 : p = 0.2,H 1 : p ≠ 0.2 p =
P( X ≥ 9) = 1 – P( X ≤ 8) or attempt critical value/region = 1 – 0.9900 = 0.01 CR X ≥ 9 A
0.01 < 0.025 or 9 ≥ 9 or 0.99 > 0.975 or 0.02 < 0.05 or lies in interval with correct interval stated. Evidence that the percentage of pupils that read Deano is not 20% A
(ii) X ~ Bin (20, 0.2) may be implied or seen in (i) or (ii) B So 0 or [9,20] make test significant. 0,9, between “their 9” and 20 B1 B1 B1 9
(b) H 0 : p = 0.2, H 1 : p ≠ 0.2 B
W ~ Bin (100, 0.2) W ~ N ( 20, 16) normal; 20 and 16 B1; B
P(X ≤ 18) = P(Z ≤ 4
- 5 − (^20) ) or 4
x − (^20) = – 1.
± cc, standardise or standardise, use z value M1 M1 A = P(Z ≤ –0.375) = 0.3520 CR X < 13.4 or 12.9 awrt 0.352 A [0.352 > 0.025 or 18 > 12.16 therefore insufficient evidence to reject H 0
Combined numbers of Deano readers suggests 20% of pupils read Deano A1 8
(c) Conclusion that they are different. B Either large sample size gives better result Or Looks as though they are not all drawn from the same population. B1 2 [19]
9. (a) Misprints are random / independent, occur singly B1, B1 2 in space and at a constant rate Context, any 2
(b) P(X = 0) = e–2.5^ M Po (2.5) = 0.08208…… = 0.0821 A1 2
(c) Y ~ Po (5) for 2 pages B Implied
P(Y > 7) = 1 – P(X ≤ 7) M Use of 1 – and correct inequality = 1 – 0.8666 = 0.1334 A1 3
(d) For 20 pages, Y ~ P 0 (50) B Y ~ N(50, 50) approx B P(Y < 40) = P( Y ≤ 39.5) M
cc ± 0.
= P
Z
standardise above M all correct A
= P (Z ≤ –1.4849) A
awrt – 1. = 1 – 0.93 = 0.07 A1 7 0. [14]
10. (a) P(R = 5) = P(R ≤ 5) – P(R ≤ 4) = 0.7216 – 0.5155 M
Can be implied = 0.2061 A1 2 Answer 0.
(OR: 5
15 C (0.3) 5 (0.7) 10 = 0.206130…)
(b) P(S = 5) = 0.2414 – 0.1321 = 0.1093 B1 1 Accept 0.1093 (AWRT) or 0.1094 (AWRT)
(OR:
- 55 e −^7.^5 = 0.10937459….)
(c) P(T = 5) = 0 B1 1 cao [4]
11. (a) Probability of success/failure is constant B Trials are independent B1 2
