2014 A-level H2 Mathematics (9740) Paper 2 Question 11 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let X and Y be the number of originals and prints sold in a week, respectively.
X \sim \mathrm{Po}(2)

Y \sim \mathrm{Po}(11)

(a)\mathrm{P}(Y > 8) = 1 - \mathrm{P}(Y \le 8) = 0.768

(b) X + Y \sim \mathrm{Po}(13)

\mathrm{P}(X + Y < 15) = 1 - \mathrm{P}(X + Y \le 14) = 0.675

(ii) Let W be the number of originals sold in n weeks.

W \sim \mathrm{Po}(2n)

\mathrm{P}(W < 3)<0.01

\mathrm{P}(W = 2) + \mathrm{P}(W = 1) + \mathrm{P}(W = 0)<0.01

e^{-2n} + e^{-2n}(2n) + e^{-2n}\frac{(2n)^{2}}{2}<0.01

e^{-2n}(1 + 2n + 2n^{2})<0.01

Using GC, smallest possible n = 5.

(iii) Since n=52 is large, by Central Limit Theorem,

B_1 + ... + B_52 \sim N(572, 572) approximately.

\mathrm{P}(B_1 + ... + B_52 > 550) = 0.821

(iv) Firstly the mean number of paintings sold per week may not be constant due to seasonal factors. Secondly, the sales might not be independent as a buyer might have preferential taste for particular artists.

Personal Comments

For (ii), students are expected to draw out the table to illustrate the choice of n. For (iii), students can either use CLT or approximate by normal distribution but do remember to do continuity correction then. I did CLT here since I haven’t done CLT for the entire paper while I did approximation in Question 7.If students do by approximation of normal distribution, they should expect the answer to be 0.816.

Comments
    pingbacks / trackbacks

    Leave a Comment

    twenty − six =

    Contact Us

    CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

    Not readable? Change text. captcha txt

    Start typing and press Enter to search