1

NORMAL DISTRIBUTION with KEY

1. A factory has a machine designed to produce 1 kg bags of sugar. It is found that the average weight of sugar in the bags is 1.02 kg. Assuming that the weights of the bags are normally distributed, find the standard

deviation if 1.7% of the bags weigh below 1 kg.

Give your answer correct to the nearest 0.1 gram. [4]

1. Let (z) = 0.017

then (–z) = 1 – 0.017 = 0.983 (M1)

z = –2.12 (A1)

But z = where x = 1 kg

Therefore = –2.12 (M1)

= 0.00943 kg = 9.4 g (to the nearest 0.1 g) (A1) (C4)

2. The random variable X is distributed normally with mean 30 and standard deviation 2.

Find p(27 X 34). [4]

2. p(27 X 34) = p (M1)

= p(–1.5 Z 2) (A1) = 0.433… + 0.477... (M1)

= 0.910 (3 sf) (accept 0.911) (A1) (C4)

3. A machine is set to produce bags of salt, whose weights are distributed normally, with a mean of 110 g and

standard deviation of 1.142 g. If the weight of a bag of salt is less than 108 g, the bag is rejected. With these

settings, 4% of the bags are rejected.

The settings of the machine are altered and it is found that 7% of the bags are rejected.

(a) (i) If the mean has not changed, find the new standard deviation, correct to three decimal

places.

The machine is adjusted to operate with this new value of the standard deviation.

(ii) Find the value, correct to two decimal places, at which the mean should be set so that only

4% of the bags are rejected.

(b) With the new settings from part (a), it is found that 80% of the bags of salt have a weight which lies

between A g and B g, where A and B are symmetric about the mean. Find the values of A and B, giving your answers correct to two decimal places.

[4]

3. Note: In all 3 parts, award (A2) for correct answers with no working. Award (M2)(A2) for correct answers with written evidence of the

correct use of a GDC (see GDC examples).

(a) (i) Let X be the random variable “the weight of a bag of salt”.

2

3034

2

3027Z

02.11

x

02.11

2

Then X ~ N(110, σ2), where σ is the new standard deviation.

Given P(X < 108) = 0.07, let Z =

Then P = 0.07 (M1)

Therefore, = –1.476 (M1)(A1)

Therefore, σ = 1.355. (A1)

GDC Example: Graphing of normal cdf with σ as the variable, and

finding the intersection with p = 0.07.

(ii) Let the new mean be µ, then X ~ N(µ, 1.3552).

Then P = 0.04 (M1)

Therefore, = 1.751 (M1)(A1)

Therefore, µ = 110.37 (A1)

GDC Example: Graphing of normal cdf with µ as the variable and finding intersection with p = 0.04.

(b) If the mean is 110.37 g then X ~ N(110.37, 1.3552), P(A < X < B) = 0.8

Then P(X < A) = 0.1, and P(X < B) = 0.9.

Therefore, = –1.282 (M1)

A = 108.63 (A1)

Therefore, = 1.282 (M1)

B = 112.11 (A1)

GDC Example: Graphing of normal cdf with X as the variable and finding intersection with p = 0.1 and 0.9.

4. The lifetime of a particular component of a solar cell is Y years, where Y is a continuous random variable with probability density function

(a) Find the probability, correct to four significant figures, that a given component fails within six months.

Each solar cell has three components which work independently and the cell will continue to run if at least

two of the components continue to work.

(b) Find the probability that a solar cell fails within six months. [7]

4. (a) Required probability

= P(Y )

= (M2)

= 0.2212. (G1)

110X

110108Z

2

355.1

108 Z

355.1

108

355.1

37.110A

355.1

37.110B

0. when e5.0

0 when 0)(

/2- y

yyf

y

21

2/1

0

2/– de5.0 yy

3

OR

Required probability = (M1)

= – (M1)

= 1 – e–1/4

= 0.2212 (4 sf) (A1)

(b) Required probability

= P(2 or 3 of the components fail in six months) (M1)

= (0.2212)2(0.7788) + (0.2212)

3 (M2)

= 0.125. (G1)

5. The diameters of discs produced by a machine are normally distributed with a mean of 10 cm and standard deviation of 0.1 cm. Find the probability of the machine producing a disc with a diameter smaller than 9.8

cm. [3]

5. Let X be the random variable “the diameter of the disc,” then X ~ N(10, 0.12).

Let Z = , the standardised normal variable (using formulae and

statistical tables).

Then P(X < 9.8) = P(Z < –2) (A1)

= 1 – (2) = 1 – 0.9773 (A1)

= 0.0227 (A1)

(using formulae and statistical tables)

OR

Using a graphic display calculator,

P(X < 9.8) = 0.0228 (A3)

Note: A different answer is obtained if a graphic display calculator is used

6. Z is the standardized normal random variable with mean 0 and variance 1. Find the value of a such that

P( Z a) = 0.75. [3]

2/1

0

2/– de5.0 yy

2/1

02/–e y

2

3

X

4

6.

From the diagram 1 – 2(1 – (a)) 0.75 (M1)

2(a) = 1.75 (A1) a = 1.15 (A1) (C3)

7. A continuous random variable X has probability density function

Find E(X). [3]

7. METHOD 1

E(X) = dx (M1)

= 0.441. (G2) (C3)

METHOD 2

E(X) = dx (M1)

= (M1)

= (ln 2) . (A1) (C3)

8. The weights of a certain species of bird are normally distributed with mean 0.8 kg and standard deviation 0.12 kg. Find the probability that the weight of a randomly chosen bird of the species lies between 0.74 kg

and 0.95 kg.

[6]

8. P(0.74 < X < 0.95) = – (M1)(A1)

elsewhere,0

,10for,)1(

4

)( 2x

xxf

1

0 2 )1(π

4

x

x

1

0 2 )1(π

4

x

x

10

2 )]1([lnπ

2x

π

2

π

4ln or

12.0

8.0–95.0

12.0

8.0–74.0

0.75

1– ( ) a

–a a

5

= (1.25) – (–0.5) (A1)

= 0.8944 – (l – 0.6915) (A1)(A1) = 0.586 (A1) (C6)

9. The probability density function f (x), of a continuous random variable X is defined by

f (x) =

Calculate the median value of X. [6]

9. Let m be the median.

Then x (4 – x2)dx = 0.5. (M1)

=> = 2 (A1)

=> [2x2 – x

4] = 2 (M1)

=> 2m2 – m

4 = 2

=> m4

– 8m2 + 8 = 0 (A1)

m = 1.08 (G2)

OR

m2 = = = 4 (4 2 ) (M1)

=> m = (A1) (C6)

Note: Award (C5) if other solutions to the equation

m4 – 8m

2 + 8 = 0 appear in the answer box.

10. The random variable X is normally distributed and

P(X 10) = 0.670

P(X 12) = 0.937.

Find E(X).

[6]

10. Let E(X) = .

From tables, z1 = 0.44, z2 = 1.53 (A1)(A1)

10 = + 0.44 (A1)

12 = + 1.53 (A1)

= (M1)

E(X) = 9.19 (A1) (C6) [6]

otherwise.,0

20),–4(4

1 2 xxx

m

04

1

xxx

m

d–4

0

3

4

1 m

0

4

1

2

32–648

2

3288 2

22–48–4

44.0–53.1

1244.0–1053.1

6

11. A random variable X is normally distributed with mean and standard deviation σ, such that

P(X > 50.32) = 0.119, and P(X < 43.56) = 0.305.

(a) Find and .

(b) Hence find P(|X – | < 5). [7]

11. (a)

(z) = 0.305 z = –0.51 (A1)

and (z) = 0.881 z = 1.18 (A1)

= 1.18 and = –0.51 (M1)

Solving simultaneously

50.32 = + 1.18σ and 43.56 = µ – 0.51σ

1.69σ = 6.76

σ = 4 µ = 45.6 (A1)(A1) 5

(b) P(X – < 5) = P(40.6 < x < 50.6) (M1) = 0.789 (G1) 2

12. The following diagram shows the probability density function for the random variable X, which is normally

distributed with mean 250 and standard deviation 50.

Find the probability represented by the shaded region.

[6]

12. z1 = 0.6 Φ(z1) = 0.7257 (A1)(A1)

z2 = –1.4 Φ(z2) = 0.0808 (A1)(A1)

Probability = 0.7257 – 0.0808 = 0.6449 = 0.645 (3 sf) (M1)(A1) (C6)

30.5%

11.9%

43.56 50.32

32.50

56.43

180 250 280

f x( )

x

7

13. Let f (x) be the probability density function for a random variable X, where

f (x)

(a) Show that k = .

(b) Calculate

(i) E(X);

(ii) the median of X. [6]

13. (a) k = 1 (M1)

k = 1 (A1)

k = (AG)

(b) (i) E(X) = (M1)

= (A1)

(ii) The median m must be a number such that

(M1)(A1)

(A1)

m3 = 4.

m = (= 1.59 to 3 sf) . (A1)

14. A continuous random variable X has probability density function given by

f (x) = k (2x – x2), for 0 x 2

f (x) = 0, elsewhere.

(a) Find the value of k.

(b) Find P(0.25 x 0.5).

[6]

otherwise,0

20for,2 xkx

8

3

2

0

2dxx

3

8

3

2

0

3 kx

8

3

)48

3(d

8

32

0

42

0

2

xxxx

2

3

22

0

2 d8

3or

2

1d

8

3

m

m

xxxx

2

10

38

3

38

3 3

0

3

mxm

2

1

8

3

m

3 4

8

14. (a) Since X is a continuous r.v. (M1)

(A1)

(A1) (C3)

(b) P(0.25 x 0.5) = (M1)

(A2) (C3)

15. Ian and Karl have been chosen to represent their countries in the Olympic discus throw. Assume that the

distance thrown by each athlete is normally distributed. The mean distance thrown by Ian in the past year

was 60.33 m with a standard deviation of 1.95 m.

(a) In the past year, 80% of Ian’s throws have been longer than x metres.

Find x, correct to two decimal places.

(b) In the past year, 80% of Karl’s throws have been longer than 56.52 m. If the mean distance of his

throws was 59.39 m, find the standard deviation of his throws, correct to two decimal places.

(c) This year, Karl’s throws have a mean of 59.50 m and a standard deviation of 3.00 m. Ian’s throws

still have a mean of 60.33 m and standard deviation 1.95 m. In a competition an athlete must have at

least one throw of 65 m or more in the first round to qualify for the final round. Each athlete is allowed three throws in the first round.

(i) Determine which of these two athletes is more likely to qualify for the final on their first

throw. (ii) Find the probability that both athletes qualify for the final. (11)

[17]

15. (a) X = length of Ian’s throw.

X N (60.33, 1.952)

P(X > x) = 0.80 z = –0.8416 (A1)

(M1)

x = 58.69 m (A1) (N3)

(b) Y = length of Karl’s throw.

Y N (59.39, 2)

P(Y > 56.52) = 0.80 z = –0.8416 (A1)

(M1)

2

0

2 1d)–2( xxxk

13

–

2

0

32

xxk

10–3

8–4

4

3k

5.0

25.0

d)( xxf

113.0256

29

95.1

33.60–8416.0–

x

39.59–52.568416.0–

9

= 3.41 (accept 3.42) (A1)

(c) (i) Y N (59.50, 3.002)

X N (60.33, 1.952)

EITHER

P(Y 65) = 0.0334 P(X 65) = 0.00831 (no (AP) here)(A2)(A2)

OR

P(Y 65) (M1)

= P(Z 1.833)

= 0.0334 (accept 0.0336) (A1)

P(X 65) (M1)

= P(Z 2.395)

= 0.0083 (accept 0.0084) (A1)

THEN

Karl is more likely to qualify since P(Y 65) P(X 65) (R1) Note: Award full marks if probabilities are not calculated but the correct conclusion is reached with the reason 1.833 < 2.395.

(ii) If p represents the probability that an athlete throws 65 metres

or more then with 3 throws the probability of qualifying for the final is

1 – (1 – p)3, or p + (1 – p)p + (1 – p)

2p,

or p3 + 3(1 – p)p

2 + 3(1 – p)

2p (M1)

Therefore P(Ian qualifies) = 1 – (1 – 0.00831)3

= 0.0247 (A1)

P (Karl qualifies) = 1 – (1 – 0.0334)3

= 0.0969 (A1)

Assuming independence (R1)

P(both qualify) = (0.0247)(0.0969) (M1)

= 0.00239 (A1) 11

Note: Depending on use of tables or gdc answers may vary from 0.00239

to 0.00244.

[17]

16. The speeds of cars at a certain point on a straight road are normally distributed with mean μ and standard

deviation σ. 15% of the cars travelled at speeds greater than 90 km h–1

and 12% of them at speeds less than

40 km h–1

. Find μ and σ. [6]

16. P(X > 90) = 0.15 and P(X < 40) = 0.12 (M1) Finding standardized values 1.036, –1.175 A1A1

00.3

50.59–65P Z

95.1

33.60–65P Z

10

Setting up the equations 1.036 = , –1.175 = (M1)

µ = 66.6, =22.6 A1A1 N2N2 [6]

17. The continuous random variable X has probability density function

f (x) = x(1 + x2) for 0 x 2,

f (x) = 0 otherwise.

(a) Sketch the graph of f for 0 x 2.

(b) Write down the mode of X.

(c) Find the mean of X.

(d) Find the median of X. [12]

17. (a)

A2

(b) Mode = 2 A1

(c) Using E(X) = (M1)

Mean= A1

= (A1)

= (1.51)

A1 N2

(d) The median m satisfies M1A1

(A1)

m4 + 2m

2 – 12 = 0

m2 = = 2.60555... (A1)

90

40

6

1

2

2

1

10 x

y

f x( )

b

axxfx d)(

2

0

42 d)(61 xxx

2

0

53

5361

xx

4568

21d)(

61

0

3 m

xxx

322

42

mm

2

4842

11

m = 1.61

A1 N3

18. The probability density function f (x) of the continuous random variable X is defined on the interval [0, a] by

Find the value of a. [6]

18. Using (M1)

(M1)(A1)

(M1)(A1)

(A1) (C6)

[6]

19. A company buys 44% of its stock of bolts from manufacturer A and the rest from manufacturer B. The

diameters of the bolts produced by each manufacturer follow a normal distribution with a standard

deviation of 0.16 mm.

The mean diameter of the bolts produced by manufacturer A is 1.56 mm.

24.2% of the bolts produced by manufacturer B have a diameter less than 1.52 mm.

(a) Find the mean diameter of the bolts produced by manufacturer B.

A bolt is chosen at random from the company’s stock.

(b) Show that the probability that the diameter is less than 1.52 mm is 0.312, to three significant figures.

(c) The diameter of the bolt is found to be less than 1.52 mm. Find the probability that the bolt was

produced by manufacturer B.

(d) Manufacturer B makes 8000 bolts in one day. It makes a profit of $1.50 on each bolt sold, on

condition that its diameter measures between 1.52 mm and 1.83 mm. Bolts whose diameters measure

less than 1.52 mm must be discarded at a loss of $0.85 per bolt.

Bolts whose diameters measure over 1.83 mm are sold at a reduced profit of $0.50 per bolt.

Find the expected profit for manufacturer B. [16]

. 3for 8

27

.3 0for 81

)(

2ax

x

xxxf

0( )d 1

a

f x x

3

0

1 9d ( 0.5625)

8 16x x

23

27d

8

a

xx

27 1 1

8 3a

27 9

8 8a

27 9 7

8 8 16a

54( 4.91)

11a

12

19. (a) Let B be the random variable “diameter of the bolts produced by

manufacturer B”.

(M1)

(A1)

(A1) N2

(b) Let A be the random variable “diameter of the bolts produced by

manufacturer A”.

(M1)

(A1)

(M1)(A1)

(3 sf) (AG) N0

(c) (M1)(A1)

(A1) N2

(d) (M1)(A1)

(M1)

(A1)

Expected gain (M1)

(A1) N2

20. A random variable X is normally distributed with mean and variance 2. If P (X 6.2) = 0.9474 and P (X

9.8) = 0.6368, calculate the value of and of .

[6]

20. P(X > 6.2) = 0.9474 gives z = 1.62 (A1)

P(X < 9.8) = 0.6368 gives z = 0.35 (A1)

(M1)(A1)

6.2 = 1.62

9.8 = 0.35

P( 1.52) 0.242B

1.52P 0.242

0.16Z

1.520.69988

0.16

1.63

1.52 1.56P( 1.52) P

0.16A Z

P( 0.25) 0.40129 (0.4013)Z

P(diameter less than1.52 mm) 0.44 0.40129 0.56 0.242

0.312

0.242 0.56

P bolt produced by B 1.520.31209

d

0.434

1.83 1.63P( 1.83) P 0.10564

0.16B Z

P(1.52 1.83) 1 0.242 0.10564B

0.65236

$ 8000 0.242 ( 0.85) 0.65236 1.50 0.10564 0.50

$ 6605.28

35.08.9

and62.12.6

σσ

13

Solving gives = 1.83 and = 9.16 (A1)(A1) (C6)

21. Use mathematical induction to prove that 5n + 9

n + 2 is divisible by 4, for n

+.

[9]

21. Let f (n) = 5n + 9

n + 2 and let Pn be the proposition that f(n) is divisible by 4.

Then f (1) = 16 A1

So P1 is true A1

Let Pn be true for n = k ie f (k) is divisible by 4 M1

Consider f (k + l) = 5k+1

+ 9k+l

+ 2 M1

= 5k(4 + 1) + 9

k(8 + 1) + 2 A1

= f (k) + 4(5k + 2 × 9

k) A1

Both terms are divisible by 4 so f (k +1) is divisible by 4. R1

Pk true Pk+l true R1

Since P1 is true, Pn is proved true by mathematical induction for n +. R1 N0