1.3 SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 14:
The area of a rectangular public parking area is 7.2 km² . Its width is 2400 m. The length, in m, of the parking area is

Solution:

\begin{array}{l} Area = Length \times Width \\ Length \times 2 .400 k m = 7 .2 k m^{2} \\ Length = \frac{7.2}{2.4} \\ = 3 k m \\ = 3000 m \\ = 3 \times 10^{3} m \end{array}

Question 15:
It is given that 30 solid metal cylinders each with a radius of 70 cm and a height of 300 cm, are melted to make 40 identical solid spheres.
Find the volume, in cm³ , of each solid sphere.

Solution:

\begin{array}{l} Volume of 40 spheres \\ = Volume of 30 cylinders \\ = 30 \times \frac{22}{7} \times 70^{2} \times 300 \\ = 138600 000 {cm}^{3} \\ Volume of 1 sphere = \frac{138600 000}{40} \\ = 3465000 {cm}^{3} \\ = 3.465 \times 10^{6} {cm}^{3} \\ = 3.47 \times 10^{6} {cm}^{3} \end{array}

8.2b Standard Normal Distribution Tables

Posted on April 23, 2020 by user

8.2b Standard Normal Distribution Tables

The characteristics of the tables of the standard normal distribution, N (0, 1) are:

Case 1

P (Z > a) is the area of the shaded region in the above diagram.

Case 2

P (–∞ < Z < ∞) = 1

Case 3

P (Z > 0) = P (Z < 0) = 0.5

Case 4

P (Z > a) = P (Z < –a)

8.2 Normal Distribution

Posted on April 23, 2020 by user

8.2 Normal Distribution

(A) Continuous Random Variable

Continuous random variable is a variable that can take any infinite value in a certain range.

(B) Normal Distribution

1. A continuous random variable, X, is normally distributed if the graph of its probability function has the following properties.

· Its curve has a bell shape and it is symmetrical at the line x = µ.

· Its curve has a maximum value at x = µ.

· The area enclosed by the normal curve and the x-axis is 1.

2. The notation of X being normally distributed with a mean, µ and a variance, σ² is X ~ N (µ, σ²).

(C) Standard Normal Distribution

If a normal random variable, X, has a mean, µ = 0 and a standard deviation, σ = 1, then X follows a standard normal distribution, i.e. X ~ N (0, 1).

(D) Curve of a Standard Normal Distribution

1. The curve of a standard normal distribution has the following properties.

· Its curve is symmetrical at the vertical line that passes through the mean, µ = 0 and has a variance, σ² = 1.

· Its curve has a maximum value at Z = 0.

· The area enclosed by the standard normal curve and the z-axis is 1.

(E) Converting a Normal Distribution to Standard Normal Distribution

A normal distribution can be converted to the standard normal distribution using the following formula:

Z = \frac{x - μ}{σ}

where,

Z = standard score or z - score

X = value of a normal random variable

µ = mean of a normal distribution

σ = standard deviation of a normal distribution

Long Questions (Question 1 & 2)

Posted on April 23, 2020 by user

Question 1:

In a school examination, 2 students out of 5 students failed Chemistry.

(a) If 6 students are chosen at random, find the probability that not more than 2 students failed Chemistry.

(b) If there are 200 Form 4 students in that school, find the mean and standard deviation of the number of students who failed Chemistry.

Solution:
(a)

\begin{array}{l} X - Number of students who failed Chemistry . \\ X ~ B (n, p) \\ X ~ B (6, \frac{2}{5}) \\ P (X = r) = {}^{n}c_{r} . p^{r} . q^{n - r} \\ P (X \leq 2) \\ = P (X = 0) + P (X = 1) + P (X = 2) \\ = {}^{6}C_{0} {(\frac{2}{5})}^{0} {(\frac{3}{5})}^{6} + {}^{6}C_{1} {(\frac{2}{5})}^{1} {(\frac{3}{5})}^{5} + {}^{6}C_{2} {(\frac{2}{5})}^{2} {(\frac{3}{5})}^{4} \\ = 0.0467 + 0.1866 + 0.3110 \\ = 0.5443 \end{array}

(b)

\begin{array}{l} X ~ B (n, p) \\ X ~ B (200, \frac{2}{5}) \\ Mean of X \\ = n p = 200 \times \frac{2}{5} = 80 \\ Standard deviation of X \\ = \sqrt{n p q} \\ = \sqrt{200 \times \frac{2}{5} \times \frac{3}{5}} \\ = \sqrt{48} \\ = 6.93 \end{array}

Question 2:

5% of the supply of mangoes received by a supermarket are rotten.

(a) If a sample of 12 mangoes is chosen at random, find the probability that at least two mangoes are rotten.

(b) Find the minimum number of mangoes that have to be chosen so that the probability of obtaining at least one rotten mango is greater than 0.85.

Solution:
(a)

X ~ B (12, 0.05)

1 – P (X ≤ 1)

= 1 – [P (X = 0) + P (X = 1)]

= 1 – [

C_{0}

(0.05)⁰ (0.95)¹² +

C_{1}

(0.05)¹ (0.95)¹¹]

= 1 – 0.8816

= 0.1184

(b)

P (X ≥ 1) > 0.85

1 – P (X = 0) > 0.85

P (X = 0) < 0.15

C_{0}

(0.05)⁰(0.95)ⁿ< 0.15

n lg 0.95 < lg 0.15

n > 36.98

n = 37

Therefore, the minimum number of mangoes that have to be chosen so that the probability of obtaining at least one rotten mango is greater than 0.85 is 37.

8.2c Probability Of An Event

Posted on April 23, 2020 by user

Example:

The masses of pears in a fruit stall are normally distributed with a mean of 220g and a variance of 100g. Find the probability that a pear that is picked at random has a mass

(a) of more than 230g.

(b) between 210g and 225g.

Hence, find the value of h such that 90% of the pears weigh more than h g.

Solution:

m= 220g

σ = √100 = 10g

Let X be the mass of a pear.

(a)

$\begin{array}{l} P (X > 230) \\ = P (Z > \frac{230 - 220}{10}) \leftarrow \begin{array}{l} Convert to standard normal \\ distribution using Z = \frac{X - μ}{σ} \end{array} \\ = P (Z > 1) \\ = 0.1587 \end{array}$

(b)

$\begin{array}{l} P (210 < X < 225) \\ = P (\frac{210 - 220}{10} < Z < \frac{225 - 220}{10}) \leftarrow \begin{array}{l} Convert to standard normal \\ distribution using Z = \frac{X - μ}{σ} \end{array} \\ = P (- 1 < Z < 0.5) \\ = 1 - P (Z > 1) - P (Z > 0.5) \\ = 1 - 0.1587 - 0.3085 \\ = 0.5328 \end{array}$

For 90% (probability = 0.9) of the pears weigh more than h g,

P(X > h) = 0.9

P(X < h) = 1 – 0.9

= 0.1

From the standard normal distribution table,

P(Z > 0.4602) = 0.1

P(Z < –0.4602) = 0.1

$\begin{array}{l} \frac{h - 220}{10} = - 0.4602 \\ h - 220 = - 4.602 \\ h = 215.4 \end{array}$

8.1a Probability Of An Event That Follows A Binomial Distribution

Posted on April 23, 2020 by user

8.1a Probability of an Event that follows a Binomial Distribution

1. For a Binomial Distribution, the probability of obtaining r numbers of successes out of n experiments is given by

P (X = r) = c_{r} . p^{r} . q^{n - r}

where

P = probability

X = discrete random variable

r = number of success (0, 1, 2, 3, …, n)

n = number of trials

p = probability of success in an experiment (0 < p <1)

q = probability of failure in an experiment (q = 1 – p)

Example 1

Kelvin has taken 3 shots in a shooting practice. The probability that Kelvin strikes the target is 0.6. X represents the number of times kelvin strikes the target.

(a) List the elements of the binomial discrete random variable X.

(b) Calculate the probability for the occurrence of each of the elements of X.

Solution:

(a) X = Number of times Kelvin strikes the target

X = {0, 1, 2, 3}

(b) X ~ B (n, p)

X ~ B (3, 0.6)

\begin{array}{l} P (X = r) = c_{r} . p^{r} . q^{n - r} \\ (i) P (X = 0) \\ = C_{0} {(0.6)}^{0} {(0.4)}^{3} \leftarrow \begin{array}{l} Probability of failure \\ = 1 - 0.6 = 0.4 \end{array} \\ = 0.064 \\ (i i) P (X = 1) \\ = C_{1} {(0.6)}^{1} {(0.4)}^{2} \\ = 0.288 \\ (i i i) P (X = 2) \\ = C_{2} {(0.6)}^{2} {(0.4)}^{1} \\ = 0.432 \\ (i v) P (X = 3) \\ = C_{3} {(0.6)}^{3} {(0.4)}^{0} \\ = 0.216 \end{array}

(c)

Long Questions (Question 5)

Posted on April 23, 2020 by user

Question 5:

The diameter of oranges harvested from a fruit orchard has a normal distribution with a mean of 3.2 cm and a variance of 2.25 cm.

Calculate

(a) the probability that an orange chosen at random from this fruit orchard has a diameter of more than 3.8 cm.

(b) the value of k if 30.5 % of the oranges have diameter less than k cm.

Solution:

µ = 3.2 cm

σ²= 2.25cm

σ = √2.25 = 1.5 cm

Let X represents the diameter of an orange.

X ~ N (3.2, 1.5²)

(a)

\begin{array}{l} P (X > 3.8) \\ = P (Z > \frac{3.8 - 3.2}{1.5}) \\ = P (Z > 0.4) \\ = 0.3446 \end{array}

(b)

\begin{array}{l} P (X < k) = 0.305 \\ P (Z < \frac{k - 3.2}{1.5}) = 0.305 \\ From the standard normal distribution table, \\ P (Z > 0.51) = 0.305 \\ P (Z < - 0.51) = 0.305 \\ \frac{k - 3.2}{1.5} = - 0.51 \\ k - 3.2 = - 0.765 \\ k = 2.435 \end{array}

Long Questions (Question 8)

Posted on April 23, 2020 by user

Question 8:
(a) A survey is carried out about red crescents in a school.
It is found that the mean of the number of red crescents is 315, the variance is 126 and the probability that a student participate in red crescents is p.
(i) Find the value of p.
(ii) If 8 students from the school are chosen at random, find the probability that more than 5 students participate in red crescents.

(b) The mass of fertiliser used in an orchard farm is normally distributed with mean 5 kg and variance 0.8 kg. Find the probability that on a particular day, more than 6 kg of fertiliser is used.

Solution:

\begin{array}{l} (a)(i) \\ n p = 315 \\ n p (1 - p) = 126 \\ 315 (1 - p) = 126 \\ 1 - p = \frac{126}{315} \\ 1 - p = 0.4 \\ p = 0.6 \\ (i i) \\ n = 8, p = 0.6 \\ P (X = r) = {}^{n}c_{r} . p^{r} . q^{n - r} \\ P (X = r) = {}^{8}C_{r} {(0.6)}^{r} {(0.4)}^{8} \\ P (X > 5) \\ = P (X = 6) + P (X = 7) + P (X = 8) \\ = {}^{8}C_{6} {(0.6)}^{6} {(0.4)}^{2} + {}^{8}C_{7} {(0.6)}^{7} {(0.4)}^{1} + {}^{8}C_{8} {(0.6)}^{8} {(0.4)}^{0} \\ = 0.20902 + 0.08958 + 0.01680 \\ = 0.3154 \end{array}

\begin{array}{l} (b) \\ P (X > 6) = P (Z > \frac{6 - 5}{\sqrt{0.8}}) \\ = P (Z > 1.12) \\ = 0.1314 \end{array}

Short Questions (Question 3 & 4)

Posted on April 23, 2020 by user

Question 3:

The masses of mangoes in a stall have a normal distribution with a mean of 200 g and a standard deviation of 30 g.

(a) Find the mass, in g, of a mango whose z-score is 0.5.

(b) If a mango is chosen at random, find the probability that the mango has a mass of at least 194 g.

Solution:

µ = 200 g

σ = 30 g

Let X be the mass of a mango.

(a)

\begin{array}{l} \frac{X - 200}{30} = 0.5 \\ X = 0.5 (30) + 200 \\ X = 215 g \end{array}

(b)

\begin{array}{l} P (X \geq 194) \\ = P (Z \geq \frac{194 - 200}{30}) \\ = P (Z \geq - 0.2) \\ = 1 - P (Z > 0.2) \\ = 1 - 0.4207 \\ = 0.5793 \end{array}

Question 4:

Diagram below shows a standard normal distribution graph.

The probability represented by the area of the shaded region is 0.3238.

(a) Find the value of k.

(b) X is a continuous random variable which is normally distributed with a mean of 80 and variance of 9.

Find the value of X when the z-score is k.

Solution:

(a)

P(Z > k) = 0.5 – 0.3238

= 0.1762

k = 0.93

(b)

µ = 80,

σ² = 9, σ = 3

\begin{array}{l} \frac{X - 80}{3} = 0.93 \\ X = 3 (0.93) + 80 \\ X = 82.79 \end{array}

8.2a Z-Score Of A Normal Distribution

Posted on April 23, 2020 by user

8.2a z–Score of a Normal Distribution

Example:

(a) A normal distribution has a mean, µ = 50 and a standard deviation σ = 10. Calculate the standard score of the value X = 35.

(b) The masses of Form 5 students of a school are normally distributed with a mean of 60 kg and a standard deviation of 15 kg.

Find

(i) the standard score of the mass of 65 kg,

(ii) the mass of a student that corresponds to the standard score of – ½.

Solution:

(a)

X ~ N (µ, σ²).

X ~ N (50, 10²)

Z = \frac{X - μ}{σ} = \frac{35 - 50}{10} = - 1.5

(b)(i)

X – Mass of a Form 5 student

X ~ N (µ, σ²).

X ~ N (60, 15²)

Z = \frac{X - μ}{σ} = \frac{65 - 60}{15} = \frac{1}{3}

Hence, the standard score of the mass of 65 kg is ⅓.

(b)(ii)

Z = – ½,

\begin{array}{l} Z = \frac{X - μ}{σ} \\ - \frac{1}{2} = \frac{X - 60}{15} \\ X - 60 = - \frac{1}{2} (15) \\ X = 52.5 \end{array}

Hence, the mass of a Form 5 student that corresponds to the standard score of – ½ is 52.5 kg.