QQQ Stats 1 Chapter 6 v1Mathematics 09 - Constraints Mathematics PDF

Preview

Summary

Mathematics 09 - Constraints MathematicsMathematics 09 - Constraints MathematicsMathematics 09 - Constraints MathematicsMathematics 09 - Constraints MathematicsMathematics 09 - Constraints MathematicsMathematics 09 - Constraints MathematicsMathematics 09 - Constraints MathematicsMathematics 09 - Con...

Description

QQQ – StatsYr1 - Chapter 6 – Statistical Distributions (v1) Total Marks: 27 (27 = Platinum, 24 = Gold, 22 = Silver, 19 = Bronze)

1.

A discrete random variable 𝑋 has the probability function 𝑘(1 + 𝑥) 𝑃(𝑋 = 𝑥) = { 0

𝑥 = 2,3,4,5 otherwise

(a) Determine the value of 𝑘. (b) Hence determine 𝑃(𝑋 ≤ 3) 2.

(3) (2)

Sue throws a fair coin 15 times and records the number of times it shows a head. (a) State the distribution to model the number of times the coin shows a head.

(2)

Find the probability that Sue records (b) exactly 8 heads, (c) at least 4 heads. (d) between 3 and 5 heads (inclusive)

(2) (2) (2)

3.

A coach company makes a profit if it books out at least 13 of its 20 coaches in a given day. The probability each coach is booked is 0.6. (a) State an assumption regarding our use of a Binomial model. (1) (b) In a week, calculate the probability that the coach company makes a profit on exactly 5 days. (3)

4.

A fair coin is tossed 4 times. Find the probability that (a) an equal number of head and tails occur (b) all the outcomes are the same, (c) the first tail occurs on the third throw.

(2) (3) (2)

5. I play a game for which the probability of winning is 0.7. If I win every game, what is the smallest number of times I play such that the probability of winning every game is less than 0.01? (3) 6. Given that 𝑋~𝐵(10,0.2), determine the smallest 𝑘 such that 𝑃(𝑋 ≥ 𝑘) ≤ 0.1

www.drfrostmaths.com

(3)

Solutions (Q2,4 © Edexcel, the rest Dr Frost) Question 1 (a) Calculates probability for at least one value of 𝑥, e.g. 𝑃(𝑋 = 2) = 𝑘(1 + 1) 3𝑘 + 4𝑘 + 5𝑘 = 12𝑘 = 1 1

𝑘 = 12

(b) 𝑃(𝑋 ≤ 3) = 3𝑘 + 4𝑘 = 7𝑘 =

A1 cso 7 12

(d) 𝑃(3 ≤ 𝑋 ≤ 5) = 𝑃(𝑋 ≤ 5) − 𝑃(𝑋 ≤ 2) = 0.15088 − 0.00369 = 0.147

3.

M1 A1

A1 cso

M1 A1

(a) Each booking is independent of each other. (Must be in context. Do not accept fixed probability or fixed number of coaches, as both are already given and therefore not assumptions) (b) Let 𝑋 be number of coaches booked in a day. 𝑋~𝐵(20,0.6) 𝑃(𝑋 ≥ 13) = 1 − 𝑃(𝑋 ≤ 12) = 0.4159 Let 𝑌 be the number of successful days in a week. 𝑌~𝐵(7,0.4159) 𝑃(𝑌 = 5) = 0.0892 (accept 0.0891)

www.drfrostmaths.com

B1

A1 B1 A1

Q5 (3 marks) 0.7𝑛 < 0.01 (𝑐𝑜𝑛𝑑𝑜𝑛𝑒 ≤) 𝑛 log 0.7 < log 0.01 (𝑀1) 𝑛 > 12.9 𝑠𝑜 𝑛 = 13 (𝐴1)

Q6) (3 marks) 𝑃(𝑋 ≥ 𝑘) ≤ 0.1 𝑃(𝑋 ≤ 𝑘 − 1) ≥ 0.9 𝑘−1= 4 𝑘=5

(𝑀1) (𝑀1) (𝐴1)

www.drfrostmaths.com

(𝑀1)...