Question

3. Buildings in a certain city centre are classified by height as tall, medium or short. The heights can be modelled by a normal distribution with mean 50 metres and standard deviation 16 metres. Buildings with a height of 70 metres are classified as tall. (i) Find the probability that a building chosen at random is classified as tall. 3 (ii) The rest of the buildings are classified as medium and short in such a way that there are twice as many medium buildings as there are short ones. Find the height below which buildings are classified as short. [6] 4. Ryan attempts a crossword puzzle every day. The number of puzzles he completes in a week (7 days) is denoted by X. (i) State two conditions that are required for X to have a binomial distribution. [2] The probability that Ryan completes a puzzle is 0.7. (ii) Use a binomial distribution to find the probability that Ryan completes at least 5 (ii) Use a binomial distribution to find the probability that, over the next 10 weeks, puzzles in a week. [4] Ryan completes 4 or fewer puzzles in exactly 3 of the 10 weeks. [4] If X ~ Bin(n, p) then P(X-x)-n Cpxg"", x 0,1,2, If x P,(a) then PX, r01,2,... Normal Approximation to the Binomial X ~ Bin(n, p) then for large n (2 30) and np, nq>5, XN (np,npq) n e2 x! If

Answer 1

3. Given that heights can be modelled by a normal distribution with mean 50 metres and standard deviation 16 metres. Buildings with a height of 70 metres are classified as tall.

(i) the probability that a building chosen at random is classified as tall is

$P(X\geq 70)=1-P(X<70)$

$P(X\geq 70)=1-P\left ( \frac{X-\mu}{\sigma }<\frac{70-50}{16} \right )$

$P(X\geq 70)=1-P\left ( z<1.25\right )$

$P(X\geq 70)=1-0.8943=0.1057$

(ii)

$P(Short)=\frac{1-0.1057}{3}=0.2981$

$\frac{X-\mu}{\sigma }=\phi ^{-1}(0.2981)$

$\frac{X-50}{16}=-0.52987$

$X=50-0.52987*16=41.52$

4.

(i) Number of trails are fixed and independent

There are two possible outcomes i.e. it may complete the puzzle or not.

(ii)

$P(X\geq 5)=\sum_{x=5}^{7}\binom{7}{x}0.7^{x}0.3^{7-x}$

$P(X\geq 5)=0.3177+0.2471+0.0824=0.6472$