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.

Short Questions (Question 1 & 2)

Posted on April 23, 2020 by user

Question 1:

Diagram below shows the graph of a binomial distribution of X.

(a) the value of h,

(b) P (X ≥ 3)

Solution:

(a)

P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

\begin{array}{l} \frac{1}{16} + \frac{1}{4} + h + \frac{1}{4} + \frac{1}{16} = 1 \\ h = 1 - \frac{5}{8} \\ h = \frac{3}{8} \end{array}

(b)

P (X ≥ 3) = P (X = 3) + P (X = 4)

P (X \geq 3) = \frac{1}{4} + \frac{1}{16} = \frac{5}{16}

Question 2:

The random variable X represents a binomial distribution with 10 trails and the probability of success is ¼.

(a) the standard deviation of the distribution,

(b) the probability that at least one trial is success.

Solution:

(a)

n = 10, p = ¼

\begin{array}{l} Standard deviation = \sqrt{n p q} \\ = \sqrt{10 \times \frac{1}{4} \times \frac{3}{4}} \\ = 1.875 \end{array}

(b)

\begin{array}{l} P (X = r) = C_{r} {(\frac{1}{4})}^{r} {(\frac{3}{4})}^{10 - r} \\ P (X \geq 1) \\ = 1 - P (X < 1) \\ = 1 - P (X = 0) \\ = 1 - C_{0} {(\frac{1}{4})}^{0} {(\frac{3}{4})}^{10} \\ = 0.9437 \end{array}

8.1b Mean, Variance And Standard Deviation Of A Binomial Distribution

Posted on April 23, 2020 by user

8.1b Mean, Variance and Standard Deviation of a Binomial Distribution

2. Determine the mean, variance and standard deviation of binomial distribution.

If X is a binomial discrete random variable such that X ~ B (n, p),

then

Mean of X,

μ = n p

Variance of X,

σ^{2} = n p q

Standard deviation of X,

σ = \sqrt{n p q}

Example 1:

A Football club organises a practice session for trainees on scoring goals from penalty kicks. Each trainee takes 8 penalty kicks. The probability that a trainee scores a goal from a penalty kick is p. After the session, it is found that the mean number of goals for a trainee is 7.2.

Find the value of p.

Solution:

Mean = np

np = 7.2

8p = 7.2

p = 0.9

Example 2:

X is a binomial random variable such that X ~ B (n, p). If its mean and standard deviation are 90 and

3 \sqrt{7}

respectively, find the value of p and of n.

Solution:

\begin{array}{l} Mean = 90 \\ n p = 90 \\ Standard deviation = 3 \sqrt{7} \\ \sqrt{n p q} = 3 \sqrt{7} \\ n p q = 9 (7) \leftarrow Square both sides \\ n p q = 63 \\ (n p) q = 63 \\ 90 q = 63 \\ q = \frac{7}{10} \\ ∴ p = 1 - \frac{7}{10} = \frac{3}{10} \\ From n p = 90, \\ n (\frac{3}{10}) = 90 \\ n = 300 \end{array}

7.1 Probability of an Event

Posted on April 23, 2020 by user

7.1 Probability of an Event

1. An experiment is a process or an action in making an observation to obtain the require results.

2. An outcome of an experiment is a possible result that can be obtained from the experiment.

3. A sample space of an experiment is the set of all the possible outcomes of an experiment.

4. The probability for the occurrence of an event A in the sample space S is

P (A) = \frac{number of outcomes of event A}{number of outcomes of sample space S}

P (A) = \frac{n (A)}{n (S)}

5. (a) The range of values of a probability is .

(b) If P (A) = 1, event A is sure to occur.

(c) If P (A) = 0, event A will not occur.

6. The complement of an event A is denoted by ̅A and the probability of a complementary event is given by

P (\bar{A}) = 1 - P (A)

Example:

A box contains 20 cards. The cards are 21 to 40 respectively. If a card is chosen at random, find the probability of obtaining

(a) an even number,

(b) an odd number greater than 29.

Solution:

The sample space, S, is

S = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40}

n (S) = 20

(a)

A = Event of obtaining an even number

A = {22, 24, 26, 28, 30, 32, 34, 36, 38, 40}

n (A) = 10

\begin{array}{l} P (A) = \frac{n (A)}{n (S)} \\ = \frac{10}{20} = \frac{1}{2} \end{array}

(b)

B = Event of obtaining an odd number greater than 29

B = {31, 33, 35, 37, 39}

n (B) = 5

\begin{array}{l} P (B) = \frac{n (B)}{n (S)} \\ = \frac{5}{20} = \frac{1}{4} \end{array}

7.3 Probability of Mutually Exclusive Events

Posted on April 23, 2020 by user

7.3 Probability of Mutually Exclusive Events

1. Two events are mutually exclusive if they cannot occur at the same time.

2. If A and B are mutually exclusive events, then

P (A υ B) = P (A) + P (B)

Example:

A bag contains 3 blue cards, 4 green cards and 5 yellow cards. A card is chosen at random from the box. Find the probability that the chosen card is green or yellow.

Solution:

Let G = event when a green card is chosen.

Y = event when a yellow card is chosen.

The sample space, S = 12, n (S) = 12

n (G) = 4 and n (Y) = 5

\begin{array}{l} P (G) = \frac{n (G)}{n (S)} = \frac{4}{12} \\ P (Y) = \frac{n (Y)}{n (S)} = \frac{5}{12} \end{array}

Events G and Y cannot occur simultaneously because we cannot obtain green card and yellow card at the same time. Therefore, events G and Y are mutually exclusive.

\begin{array}{l} G \cap Y = \emptyset \\ P (G \cup Y) = P (G) + P (Y) \\ = \frac{4}{12} + \frac{5}{12} \\ = \frac{9}{12} = \frac{3}{4} \end{array}

7.4 Probability of Independent Events

Posted on April 23, 2020 by user

7.4 Probability of Independent Events

1. In an experiment, if the outcomes of event A do not influence the outcomes of event B, then the two events are independent.

2. If A and B are two independent events, the probability for the occurrence of events A and B is

P (A ∩ B) = P (A) × P (B)

3. The concept of the probability of two independent events can be expanded to three or more independent events. If A, B and C are three independent events, the probability for the occurrence of events A, B and C is

P (A ∩ B ∩ C) = P (A) x P (B) x P (C)

4. A tree diagram can be constructed to show all the possible outcomes of an experiment.

Example:

Fatimah, Emily and Rani are to take a written driving test. The probability that they pass the test are ½, ⅔ and ¾ respectively. Calculate the probability that

(a) only one of them passes the exam,

(b) at least two of them pass the exam,

Solution:

Let P = Pass and F = Fail

The tree diagram is as follows.

(a)

P (only one of them passes the exam)

= P(PFF or FPF or FFP)

= P(PFF) + P(FPF) + P(FFP)

\begin{array}{l} = \frac{1}{24} + \frac{1}{12} + \frac{1}{8} \\ = \frac{1}{4} \end{array}

(b)

P (at least two of them pass the exam)

= P (PPP or PPF or PFP or FPP)

= P (PPP) + P (PPF) + P (PFP) + P (FPP)

\begin{array}{l} = \frac{1}{4} + \frac{1}{12} + \frac{1}{8} + \frac{1}{4} \\ = \frac{17}{24} \end{array}

(c)

P (at least one of them passes the exam)

= 1 – P (all of them fail)

= 1 – P (FFF)

\begin{array}{l} = 1 - \frac{1}{24} \\ = \frac{23}{24} \end{array}

Long Questions (Question 1)

Posted on April 23, 2020 by user

Question 1:

Peter, William and Roger compete with each other in shooting a target. The probabilities that they strike the target are

\frac{2}{5}, \frac{3}{4} and \frac{2}{3}

respectively. Calculate the probability that

(a) all the three of them strike the target,

(b) only one of them strikes the target,

Solution:

Let S = Strike and M = Missed

\begin{array}{l} Probability of Peter missed the target = \frac{3}{5} \\ Probability of William missed the target = \frac{1}{4} \\ Probability of Roger missed the target = \frac{1}{3} \end{array}

(a)

\begin{array}{l} Probability (all three persons strike the target) \\ = \frac{2}{5} \times \frac{3}{4} \times \frac{2}{3} \\ = \frac{1}{5} \end{array}

(b)

\begin{array}{l} Probability (only one of them strikes the target) \\ = P (only Peter struck) + P (only William struck) + P (only Roger struck) \\ = (\frac{2}{5} \times \frac{1}{4} \times \frac{1}{3}) + (\frac{3}{5} \times \frac{3}{4} \times \frac{1}{3}) + (\frac{3}{5} \times \frac{1}{4} \times \frac{2}{3}) \\ = \frac{1}{30} + \frac{3}{20} + \frac{1}{10} \\ = \frac{17}{60} \end{array}

(c)

\begin{array}{l} Probability (at least one of them strikes the target) \\ = 1 - P (all missed the target) \\ = 1 - (\frac{3}{5} \times \frac{1}{4} \times \frac{1}{3}) \\ = 1 - \frac{1}{20} \\ = \frac{19}{20} \end{array}