# IB Math Analysis & Approaches Question bank-Topic 5.9 normal distribution SL Paper 1

## Question

Let X be normally distributed with mean 100 cm and standard deviation 5 cm.

On the diagram below, shade the region representing $${\rm{P}}(X > 105)$$ . a.

Given that $${\rm{P}}(X < d) = {\rm{P}}(X > 105)$$ , find the value of $$d$$ .


b.

Given that $${\rm{P}}(X > 105) = 0.16$$ (correct to two significant figures), find $${\rm{P}}(d < X < 105)$$ .


c.

## Markscheme A1A1     N2

Note: Award A1 for vertical line to right of mean, A1 for shading to right of their vertical line.

a.

evidence of recognizing symmetry     (M1)

e.g. $$105$$ is one standard deviation above the mean so $$d$$ is one standard deviation below the mean, shading the corresponding part, $$105 – 100 = 100 – d$$

$$d = 95$$     A1     N2

[2 marks]

b.

evidence of using complement     (M1)

e.g. $$1 – 0.32$$ , $$1 – p$$

$${\rm{P}}(d < X < 105) = 0.68$$     A1     N2

[2 marks]

c.

MAA SL 4.9-4.10 DISCRETE DISTRIBUTIONS – BINOMIAL [concise]-neha

### Question

[with / without GDC]

The probability distribution of the discrete random variable X is given by the table Find

(a) the expected value $$E(X)$$ of $$X$$.

(b) the mode of $$X$$.

(c) the median of $$X$$.

(d) the lower quartile $$Q_{1}$$ and the upper quartile $$Q_{3}$$.

Ans

(a) $$2.35$$       (b) mode =$$1$$         (c) median = $$2$$            (d) $$Q_{1}$$ = $$1$$         $$Q_{3}$$ = $$3.5$$

### Question

[without GDC]

The probability distribution of the discrete random variable X is given by the table Find the values of $$a$$ and $$b$$ given that $$E(X)$$ = $$2.2$$

Ans

a = $$0.2$$         b = $$0.4$$

### Question

[without GDC]

The probability distribution of the discrete random variable X is given by the table. Nikos selects a number at random.

If he selects $$1$$ he earns $$10$$ €. If he selects $$2$$ he earns $$5$$ €. If he selects $$3$$ he loses $$4$$ €

(a) Find the expected value of $$X$$.

(b) Find the expected value of the profit for Nikos. Is the game fair?

Ans

$$E(X)$$ = $$2.2$$\)       E(Profit) = $$0$$,  so it is fair.

### Question

[without GDC]

A discrete random variable X has its probability distribution given by

$$P(X=x)=\frac{x}{6}$$  where x is $$1, 2, 3$$.

(a)  Complete the following table showing the probability distribution for $$X$$. (b)  Find $$E(X)$$.

Ans

(a) (b)    Find $$E(X)=7/3$$

### Question

[without GDC]

Each of the following 10 words is placed on a card and put in a hat.

ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN

We pick a card at random. Let $$X$$ be the size (number of letters) of the corresponding word.

(a) Give the probability distribution for $$X$$ (i.e. the table of probabilities)

(b) Find the expected number of $$X$$.

Ans

(a) (b)  Find $$E(X) = 3.9$$

### Question

[with GDC]

The probability distribution of the discrete random variable$$X$$ is given by the following table. (a) Find the value of $$p$$.

(b) Calculate the expected value of $$X$$.

Ans

(a) $$(0.4 + p + 0.2 + 0.07 + 0.02 = 1), \Rightarrow p = 0.31$$

(b) $$E(X) = 1(0.4) + 2(0.31) + 3(0.2) + 4(0.07) + 5(0.02) = 2$$

### Question

[without GDC]

A discrete random variable X has a probability distribution as shown in the table below. (a)  Find the value of $$a + b$$.

(b)  Given that $$E(X) =1.5$$, find the value of $$a$$ and of $$b$$.

Ans

(a) $$0.1 + a + 0.3 + b = 1 \Rightarrow a + b = 0.6$$

(b) $$0\times 0.1+1\times a+2\times 0.3+3\times b$$

$$0 + a + 0.6 + 3b = 1.5$$

$$a + 3b = 0.9$$

Solving simultaneously gives

$$a = 0.45$$        $$b = 0.15$$

### Question

[with GDC]

The following table shows the probability distribution of a discrete random variable $$X$$. (a) Find the value of $$k$$.

(b) Find the expected value of $$X$$.

Ans

(a)  $$10k^{2}+3k+0.6=1\Rightarrow 10k^{2}+3k+0.4=0\Leftrightarrow k=0.1$$     (by GDC)

(b)  $$E(X)=-1\times 0.2+2\times 0.4+3\times 0.3=1.5$$

### Question

[without GDC]

A discrete random variable $$X$$ has its probability distribution given by

$$P(X = x) = k(x + 1)$$, where $$x$$ is $$0, 1, 2, 3, 4$$.

(a)  Complete the following table showing the probability distribution for $$X$$ (in terms of $$k$$) (b)  Show that  $$k=\frac{1}{15}$$

(c)  Find $$E(X)$$.

Ans

(a) (b)     $$k\times 1+k\times 2+k\times 3+k\times 4+k\times 5=15k=1\Leftrightarrow k=\frac{1}{15}$$

(c)      $$E(X)=0\times \frac{1}{15}+1\times \frac{2}{15}+2\times \frac{3}{15}+3\times \frac{4}{15}+4\times \frac{5}{15}=\frac{40}{15}=\frac{8}{3}$$

### Question

[without GDC]

Three students, Kim, Ching Li and Jonathan each have a pack of cards, from which they select a card at random. Each card has $$a 0, 3, 4$$, or $$9$$ printed on it.

(a)     Kim states that the probability distribution for her pack of cards is as follows. Explain why Kim is incorrect.

(b)    Ching Li correctly states that the probability distribution for her pack of cards is as follows. Find the value of $$k$$.

(c)     Jonathan correctly states that the probability distribution for his pack of cards is given by $$P(X=x)=\frac{x+1} {20}$$. One card is drawn at random from his pack. Calculate the probability that the number on the card drawn

(i)  is $$0$$.        (ii) is greater than $$0$$.

Ans

(a)     Sum = $$1.3$$ which is greater than $$1$$

(b)     3k + 0.7 = 1 \Rightarrow k = 0.1

(c)    (i)    $$P(X=0)=\frac{0+1}{20}=\frac{1}{20}$$

(ii)   $$P(X>0)=1-P(X=0)=\frac{19}{20} \left ( or \frac{4}{20}+\frac{5}{20}+\frac{10}{20} \right )=\frac{19}{20}$$

Notice: in fact, the probability distribution is ### Question

[with GDC]

A fair coin is tossed eight times.

(a)  Calculate the probability of obtaining (b)  Find the expected number of heads and the variance of the number of heads Ans

(a) (b) ### Question

[without GDC]

The random variable $$X$$ follows the binomial distribution B($$n,p$$). Given that $$E(X)$$ = $$10$$ and Var($$X$$) = $$6$$ find the values of $$n$$ and $$p$$.

Ans

$$np=10$$ and $$np(1-p)=6$$. Hence $$10(1-p)=6 \Leftrightarrow \(p=0.4$$ and $$n=25$$

### Question

[with GDC]

A fair coin is tossed five times. Calculate the probability of obtaining

Ans

B$$(n,p)$$ with $$n= 5$$ and $$p= \frac{1}{2}$$

(a) $$P(X = 3) = 0.3125 … = 0.313$$

(b) $$P(X \geq 1) = 0.969$$

### Question

[with GDC]

The probability of obtaining heads on a biased coin is $$0.18$$. The coin is tossed seven times.

(a) Find the probability of obtaining exactly two heads.

(b) Find the probability of obtaining at least two heads.

Ans

$$B( n, p)$$   with   $$n = 7$$  and  $$p = 0.18$$

(a) $$P(X = 2) = 0.252$$

(b) $$P(X \geq 2) = 0.368$$

### Question

[with GDC]

A factory makes switches. The probability that a switch is defective is $$0.04$$.

The factory tests a random sample of $$100$$ switches.

(a) Find the mean number of defective switches in the sample.

(b) Find the probability that there are exactly six defective switches in the sample.

(c) Find the probability that there is at least one defective switch in the sample.

Ans

$$B( n,p)$$  with    $$n = 100$$   and   $$p = 0.04$$

(a) mean = $$np = 100$$ \times $$0.04 = 4$$

(b) $$P(X = 6)$$ = $$0.105$$

(c) $$P(X \geq 1) = 0.983$$

### Question

[with GDC]

A factory makes calculators. Over a long period, $$2$$% of them are found to be faulty. A random sample of $$100$$ calculators is tested.

(a) Write down the expected number of faulty calculators in the sample.

(b) Find the probability that three calculators are faulty.

(c) Find the probability that more than one calculator is faulty.

Ans

$$X \sim B(100,0.02)$$

(a) $$E(X) = 100 \times 0.02 = 2$$

(b) (i) $$P(X = 3) = 0.182$$                 (ii) $$P(X > 1) = 0.597$$

### Question

[with GDC]

A box contains $$35$$ red discs and $$5$$ black discs. A disc is selected at random and its colour noted. The disc is then replaced in the box.

(a)      In eight such selections, what is the probability that a black disc is selected

(i) exactly once?

(ii) at least once?

(b)     The process of selecting and replacing is carried out $$400$$ times.

What is the expected number of black discs that would be drawn?

Ans

$$p$$(Red)=$$\frac{35}{40}=\frac{7}{8}$$                            $$p$$(Black)=$$\frac{5}{40}=\frac{1}{8}$$

(a) B( n, p ) with n = 5, p=$$\frac{1}{8}$$

(i) $$p$$(one black) = P($$X = 1) = 0.393$$ to $$3$$ s.f.

(ii) $$p$$(at least one black) = $$P(X\geq 1)=0.656$$

(b) 400 draws: expected number of blacks =$$\frac{400}{8}=50$$

### Question

[with GDC]

The probability of obtaining heads on a biased coin is $$\frac{1}{3}$$

(a) Sammy tosses the coin three times. Find the probability of getting

(ii) two heads and one tail.

(b) Amir plays a game in which he tosses the coin $$12$$ times.

(i) Find the expected number of heads.

(ii) Amir wins $$10$$ for each head obtained and loses $$6$$ for each tail. Find his expected winnings.

Ans

(a)     B$$(n, p)$$ with $$n=3, p=\frac{1}{3}$$

(i) $$P(X = 3) = 0.0370$$  or  P($$3$$H)= $$\left ( \frac{1}{3}^{3} \right )=\frac{1}{27}$$

(ii) $$P(X = 3) = 0.222$$  or P(2H, 1T) = $$3\left ( \frac{1}{3}^{2} \right )\frac{2}{3}=\frac{2}{9}$$

(b)     (i) expected number of heads = $$np$$ = $$\left ( \frac{1}{3}\times 12 \right )=4$$

(ii)     $$4$$ heads, so $$8$$ tails

E(winnings) = $$4\times 10-8\times 6(=40-48)$$= $$– 8$$

### Question

[without GDC]

Two fair four-sided dice, one red and one green, are thrown. For each die, the faces are labelled 1, 2, 3, 4. The score for each die is the number which lands face down.

The sample space is shown below: (a)  Write down the probability that two scores of $$4$$ are obtained.

Let $$X$$ be the number of $$4$$s that land face down.

(b)  Complete the following probability distribution for $$X$$. (c) Find  $$E(X)$$.

Chris plays a game where he rolls the dice above. If two $$4$$s are obtained he wins $$20$$€.

If only one $$4$$ is obtained he wins $$5$$€. If no $$4$$ is obtained he loses $$2$$€

(d)  Find the expected amount earned in one game.

(e)  If Chris plays this game $$100$$ times find the amount he is expected to win.

Ans

(a)    Probability of two $$4$$s is  $$\frac{1}{16}(=0.0625)$$

(b) (c)     $$E(X)=\sum_{0}^{2} xP(X=x),=0\times \frac{9}{16}+1\times \frac{6}{16}+2\times \frac{1}{16}\frac{8}{16}\left ( =\frac{1}{2} \right )$$

or $$E(X)=np=2\times \frac{1}{4}=\frac{1}{2}$$

(d) Expected amount = $$-2\times \frac{9}{16}+5\times \frac{6}{16}+20\times \frac{1}{16}= 2$$€

(e)    $$100\times 2$$= $$200$$ €

### Question

[with GDC]

Bag A contains $$2$$ red balls and $$3$$ green balls. Two balls are chosen at random from the bag without replacement. Let $$X$$ denote the number of red balls chosen. The following

table shows the probability distribution for $$X$$ (a) Calculate E(X), the mean number of red balls chosen.

Bag B contains $$4$$ red balls and $$2$$ green balls. Two balls are chosen at random from bag B.

(b)     (i) Draw a tree diagram to represent the above information, including the probability of each event.

(ii) Hence find the probability distribution for $$Y$$, where $$Y$$ is the number of red balls chosen.

A standard die with six faces is rolled. If a $$1$$ or $$6$$ is obtained, two balls are chosen from bag A, otherwise two balls are chosen from bag B.

(c) Calculate the probability that two red balls are chosen.

(d) Given that two red balls are obtained, find the conditional probability that a $$1$$ or $$6$$ was rolled on the die.

Ans

(a)    $$E(X)=0\times \frac{3}{10}+1\times \frac{6}{10}+2\times \frac{1}{10}=\frac{8}{10} (0.8)$$

(b)     (i) (ii)     $$P(Y=0)=\frac{2}{5}\times \frac{1}{5}=\frac{2}{30}$$

$$P(Y=1)= P(RG)+P(GR \left ( =\frac{4}{6}\times \frac{2}{5}+\frac{2}{6}\times \frac{4}{5} \right )=\frac{16}{30}$$

$$P(Y=2)= \frac{4}{6}\times \frac{3}{5}=\frac{12}{30}$$

Forming a distribution (c)    $$P(RR)=\frac{1}{3}\times \frac{1}{10}+\frac{2}{3}\times \frac{12}{30}=\frac{27}{90} \left ( \frac{3}{10}, 0.3 \right )$$

(d)    $$P(1 or 6|RR)= P(A|RR)= \frac{P(A\cap RR)}{P(RR)}=\frac{1}{30}\div \frac{27}{90}=\frac{3}{27} \left ( \frac{1}{9},0.111 \right )$$

### Question

[without GDC]

A four-sided die has three blue faces and one red face. The die is rolled. Let $$B$$ be the event a blue face lands down, and $$R$$ be the event a red face lands down.

(a)     Write down the values of

(i) $$P(B)$$

(ii) $$P(R)$$

(b)     If the blue face lands down, the die is not rolled again. If the red face lands down, the die is rolled once again. This is represented by the following tree diagram, where $$p, s, t$$ are

probabilities. Find the value of $$p$$, of $$s$$ and of $$t$$.

Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores $$2$$ and is finished. If the red face lands down, he scores $$1$$ and rolls one more time. Let $$X$$ be the total score obtained.

(c)      (i) Show that $$P(X=3)=\frac{3}{16}$$.

(ii) Find P $$(X = 2)$$.

(d)     (i) Construct a probability distribution table for $$X$$.

(ii) Calculate the expected value of $$X$$.

(e)      If the total score is $$3$$, Guiseppi wins $$10$$. If the total score is $$2$$, Guiseppi gets nothing. He plays the game twice. Find the probability that he wins exactly $$10$$.

Ans

(a)       (i)    P$$(B)=\frac{3}{4}$$             (ii)  P$$(R)=\frac{1}{4}$$

(b)      $$p=\frac{3}{4}\ s=\frac{1}{4},\ t=\frac{3}{4}$$

(c)      (i)    $$P(X=3)$$=P(getting $$1$$ and $$2$$)=$$\frac{1}{4}\times \frac{3}{4}=\frac{3}{16}$$

(ii)    $$P(X=2)=\frac{1}{4}\times \frac{1}{4}+\frac{3}{4} \left ( or\ 1-\frac{3}{16} \right )= \frac{13}{16}$$

(d)     (i) (ii)    $$E(X)= 2\left ( \frac{13}{16} \right )+3\left ( \frac{3}{16} \right )=\frac{35}{16}$$

(e)      win $$10 \Rightarrow$$ scores $$3$$ one time, $$2$$ other time

P(win $$10$$) = $$P(3)\times P(2)+P(2)\times P(3)=2\left ( \frac{13}{16}\times \frac{3}{16} \right ), =\frac{78}{256}\left ( =\frac{39}{128} \right )$$

### Question

[with GDC]

A pair of fair dice is thrown.

(a) Complete the tree diagram below, which shows the possible outcomes. Let $$E$$ be the event that exactly one four occurs when the pair of dice is thrown.

(b) Calculate $$P(E)$$.

The pair of dice is now thrown five times.

(c) Calculate the probability that event $$E$$ occurs exactly three times in the five throws.

(d) Calculate the probability that event $$E$$ occurs at least three times in the five throws.

(a) (b)      $$P(E)=\frac{1}{6}\times \frac{5}{6}+\frac{5}{6}\times \frac{1}{6} \left ( =\frac{5}{36}+\frac{5}{36} \right )=\frac{10}{36} \left ( \frac{5}{18}\ or\ 0.0278\right )$$
(c)      $$X\sim B \left ( 5, \frac{5}{18} \right )$$
P$$(X=3)=0.112$$                                                                                                            [in fact $$\binom{5}{3}\left ( \frac{5}{3} \right )^{3}\left ( \frac{13}{8} \right )^{2}=0.112$$]
(d)     $$P(X\geq 3)=0.135$$