Class 12- Applied Mathematics-NCERT Handbook Solutions-Unit-4-Probability Distribution -

NCERT Handbook Solutions

Class 12-Applied Mathematics

4.8 CHECK YOUR PROGRESS

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
a.

X	-1	0	1	2
P(X)	0.1	0.8	0.001	0.2

X	1	2	3	4	5
P(X)	0.1	0.4	0.05	-0.2	0.2

X	-2	2	5
P(X)	0.5	0.2	0.3

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Answer:

a. Not a valid distribution.
Reason: The sum of probabilities must be \( 1 \).
\( \sum P(X) = 0.1 + 0.8 + 0.001 + 0.2 = 1.101 \ne 1 \).
b. Not a valid distribution.
Reason: Probability cannot be negative. \( P(X=4) = -0.2 \), which is invalid.
c. Valid distribution.
Reason: All \( P(X) \ge 0 \) and \( \sum P(X) = 0.5 + 0.2 + 0.3 = 1 \).

A lady's bag contains \( 2 \) black and \( 1 \) red pens. One pen is drawn at random and then put back in the box after noting its colour. The process is repeated again. If \( X \) denotes the number of red pens recorded in the two draws. Describe \( X \).

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Answer:

The random variable \( X \) represents the number of red pens in two draws with replacement.

Possibilities:

Total pens = \( 3 \) (\( 2 \) Black, \( 1 \) Red).
Probability of Red (\( R \)) = \( 1/3 \).
Probability of Black (\( B \)) = \( 2/3 \).

Possible values for \( X \) are \( 0, 1, 2 \).

\( P(X=0) \) (No Red, i.e., BB): \[ \frac{2}{3} \times \frac{2}{3} = \frac{4}{9} \]
\( P(X=1) \) (One Red, i.e., BR or RB): \[ \left(\frac{2}{3} \times \frac{1}{3}\right) + \left(\frac{1}{3} \times \frac{2}{3}\right) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9} \]
\( P(X=2) \) (Two Reds, i.e., RR): \[ \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \]

Probability Distribution:

\( X \)	0	1	2
\( P(X) \)	\( 4/9 \)	\( 4/9 \)	\( 1/9 \)

What is the mean of the numbers obtained on throwing a die having written \( 1 \) on three faces, \( 2 \) on two faces and \( 5 \) on one face?

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Answer: \( 2 \)

Given distribution on the die:

Number \( 1 \): on 3 faces \( \Rightarrow P(1) = 3/6 = 1/2 \)
Number \( 2 \): on 2 faces \( \Rightarrow P(2) = 2/6 = 1/3 \)
Number \( 5 \): on 1 face \( \Rightarrow P(5) = 1/6 \)

Mean (Expected Value) \( E(X) = \sum X \cdot P(X) \):

\[ E(X) = 1 \cdot \frac{3}{6} + 2 \cdot \frac{2}{6} + 5 \cdot \frac{1}{6} \] \[ E(X) = \frac{3}{6} + \frac{4}{6} + \frac{5}{6} = \frac{12}{6} = 2 \]

Raheem tossed a fair coin \( 10 \) times, find the probability of (i) exactly six heads (ii) at least six heads (iii) at most six heads.

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Answer:

This is a Binomial Distribution problem with \( n=10, p=0.5, q=0.5 \).

Formula: \( P(X=r) = \binom{n}{r} p^r q^{n-r} = \binom{10}{r} (0.5)^{10} \).

(i) Exactly six heads (\( r=6 \)):

\[ P(X=6) = \binom{10}{6} (0.5)^{10} = \frac{210}{1024} = \frac{105}{512} \approx 0.205 \]

(ii) At least six heads (\( X \ge 6 \)):

\[ P(X \ge 6) = P(6) + P(7) + P(8) + P(9) + P(10) \] \[ = \frac{1}{1024} [ \binom{10}{6} + \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} ] \] \[ = \frac{1}{1024} [ 210 + 120 + 45 + 10 + 1 ] = \frac{386}{1024} = \frac{193}{512} \approx 0.377 \]

(iii) At most six heads (\( X \le 6 \)):

\[ P(X \le 6) = 1 - P(X > 6) = 1 - [P(7) + P(8) + P(9) + P(10)] \] \[ = 1 - \frac{176}{1024} = \frac{848}{1024} = \frac{53}{64} \approx 0.828 \]

Find the probability of getting \( 5 \) exactly twice in \( 7 \) throws of a fair die.

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Answer:

Given:

\( n = 7 \)
Probability of getting '5' (\( p \)) = \( 1/6 \)
Probability of not getting '5' (\( q \)) = \( 5/6 \)
Successes required (\( r \)) = \( 2 \)

Using Binomial formula \( P(X=r) = \binom{n}{r} p^r q^{n-r} \):

\[ P(X=2) = \binom{7}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5 \] \[ = 21 \cdot \frac{1}{36} \cdot \left(\frac{5}{6}\right)^5 \] \[ = \frac{7}{12} \left(\frac{5}{6}\right)^5 \]

Let \( X \) denote the number of hours a class XII student studies during a randomly selected school day. The probability that \( X \) can take the values \( x_i \), for an unknown constant \( k \) is given by: \[ P(X=x_i) = \begin{cases} 0.1, & \text{if } x_i = 0 \\ kx_i, & \text{if } x_i = 1 \text{ or } 2 \\ k(5-x_i), & \text{if } x_i = 3, 4 \end{cases} \] a. Find the value of \( k \).
b. What is the probability that the student studied for at least two hours? Exactly two hours? At most two hours?

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Answer: a. \( k=0.15 \); b. \( 0.75, 0.3, 0.55 \)

a. Find \( k \):

Sum of probabilities must be \( 1 \):

\[ P(0) + P(1) + P(2) + P(3) + P(4) = 1 \]

\( P(0) = 0.1 \)
\( P(1) = 1k \)
\( P(2) = 2k \)
\( P(3) = k(5-3) = 2k \)
\( P(4) = k(5-4) = k \)

\[ 0.1 + k + 2k + 2k + k = 1 \] \[ 0.1 + 6k = 1 \Rightarrow 6k = 0.9 \Rightarrow k = 0.15 \]

b. Probabilities:

At least two hours (\( X \ge 2 \)): \[ P(2) + P(3) + P(4) = 2k + 2k + k = 5k = 5(0.15) = 0.75 \]
Exactly two hours (\( X = 2 \)): \[ P(2) = 2k = 2(0.15) = 0.3 \]
At most two hours (\( X \le 2 \)): \[ P(0) + P(1) + P(2) = 0.1 + k + 2k = 0.1 + 3(0.15) = 0.1 + 0.45 = 0.55 \]

How many times must Sumit toss a fair coin so that the probability of getting at least one head is more than \( 90\% \)?

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Answer: \( 4 \) times

Let \( n \) be the number of tosses.

Probability of getting at least one head \( = 1 - P(\text{No heads}) \).

\( P(\text{No heads}) = (1/2)^n \).

We want:

\[ 1 - \left(\frac{1}{2}\right)^n > 0.90 \] \[ 0.10 > \left(\frac{1}{2}\right)^n \] \[ \frac{1}{10} > \frac{1}{2^n} \Rightarrow 2^n > 10 \]

Powers of 2:

\( 2^3 = 8 \) (Not greater than 10)
\( 2^4 = 16 \) (Greater than 10)

Therefore, minimum tosses \( n = 4 \).

A pair of dice is thrown and the random variable \( X \) represents the sum of the numbers that appear on the two dice. Calculate the mathematical expectation of \( X \).

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Answer: \( 7 \)

The expected value of the sum of two dice is the sum of their individual expected values.

\( E(X) = E(D_1 + D_2) = E(D_1) + E(D_2) \).

Expected value of a single die:

\[ E(D) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5 \] \[ E(X) = 3.5 + 3.5 = 7 \]

Find the variance of a bernoulli random variable whose probability of success is \( 0.6 \).

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Answer: \( 0.24 \)

For a Bernoulli trial:

\( p = 0.6 \)
\( q = 1 - p = 0.4 \)

Variance \( = p \times q \).

\[ \text{Variance} = 0.6 \times 0.4 = 0.24 \]

Q10

If the mean and variance of a binomial distribution are \( 4/3 \) and \( 8/9 \) resp. find \( P(X=1) \).

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Answer: \( 32/81 \)

Given:

Mean \( np = 4/3 \)
Variance \( npq = 8/9 \)

Divide Variance by Mean to find \( q \):

\[ q = \frac{npq}{np} = \frac{8/9}{4/3} = \frac{8}{9} \times \frac{3}{4} = \frac{2}{3} \]

Since \( p + q = 1 \), \( p = 1 - 2/3 = 1/3 \).

Find \( n \):

\[ n(1/3) = 4/3 \Rightarrow n = 4 \]

Now find \( P(X=1) \) using Binomial formula:

\[ P(X=1) = \binom{4}{1} (1/3)^1 (2/3)^3 \] \[ = 4 \times \frac{1}{3} \times \frac{8}{27} = \frac{32}{81} \]

Q11

What is the expected value of number of tails on a throw of a fair coin?

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Answer: \( 0.5 \)

For a single coin toss, the possible outcomes for number of tails (\( X \)) are \( 0 \) (Head) and \( 1 \) (Tail).

\( P(X=0) = 0.5 \)
\( P(X=1) = 0.5 \)

\[ E(X) = 0 \times 0.5 + 1 \times 0.5 = 0.5 \]

Q12

A customer care company receives an average of \( 4.5 \) calls every \( 5 \) minutes. If each customer executive can handle one of these calls over the \( 5 \)-minute period. But if an executive is not unavailable to take the call, then the call is put on hold. Assuming that the calls received by the customer care company follows a Poisson distribution, what is the minimum number of customer executives are needed on duty so that calls received are placed on hold for at the most \( 10\% \) of the time?

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Answer: \( 7 \) Executives

Given:

Mean \( \lambda = 4.5 \)
We need to find the number of executives \( k \) such that the probability of calls exceeding \( k \) is at most \( 10\% \).
Mathematically: \( P(X > k) \le 0.10 \) or \( P(X \le k) \ge 0.90 \).

Using Cumulative Poisson Probability table or calculation for \( \lambda = 4.5 \):

\( P(X \le 5) \approx 0.7029 \)
\( P(X \le 6) \approx 0.8311 \)
\( P(X \le 7) \approx 0.9134 \)

Since \( P(X \le 7) \) is the first value greater than \( 0.90 \), we need \( 7 \) executives to ensure that more than \( 7 \) calls (overflow/hold) happen less than \( 10\% \) of the time.

Q13

A statistician records the number of trucks approaching a particular intersection to analyze the flow of traffic. He observes that on an average \( 1.6 \) trucks approach the intersection every minute. Assuming that the number of trucks approaching the intersection, follow a Poisson distribution model, what is the probability that \( 3 \) or more trucks will approach the intersection within a minute?

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Answer: \( 0.2166 \)

Given:

Average rate of trucks per minute, \( \lambda = 1.6 \)
The distribution follows a Poisson distribution.

The probability mass function for a Poisson distribution is given by:

\[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \]

We need to find the probability that \( 3 \) or more trucks will approach the intersection, i.e., \( P(X \ge 3) \).

Since the total probability is \( 1 \):

\[ P(X \ge 3) = 1 - P(X < 3) \] \[ P(X \ge 3) = 1 - [P(X=0) + P(X=1) + P(X=2)] \]

Calculating each term using \( \lambda = 1.6 \) and \( e^{-1.6} \approx 0.2019 \):

For \( x = 0 \): \[ P(X=0) = \frac{e^{-1.6} (1.6)^0}{0!} = e^{-1.6} \approx 0.2019 \]
For \( x = 1 \): \[ P(X=1) = \frac{e^{-1.6} (1.6)^1}{1!} = 1.6 \times e^{-1.6} \approx 1.6 \times 0.2019 = 0.3230 \]
For \( x = 2 \): \[ P(X=2) = \frac{e^{-1.6} (1.6)^2}{2!} = \frac{2.56}{2} \times e^{-1.6} = 1.28 \times 0.2019 \approx 0.2584 \]

Summing these probabilities:

\[ P(X < 3) = 0.2019 + 0.3230 + 0.2584 = 0.7833 \]

Now, calculate \( P(X \ge 3) \):

\[ P(X \ge 3) = 1 - 0.7833 = 0.2167 \]

Therefore, the probability is approximately \( 0.2167 \).

Q14

A computer disk manufacturer tests disk quality on random basis before approving it. The approval is based on the number of errors in a test area on each disk and follows Poisson distribution with \( \lambda = 0.2 \). What is the percentage of test areas having two or a smaller number of errors?

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Answer: \( 99.9\% \)

Given:

Mean number of errors, \( \lambda = 0.2 \)
We need to find the percentage of areas with \( X \le 2 \).

Formula:

\[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \]

We calculate \( P(X \le 2) = P(X=0) + P(X=1) + P(X=2) \).

Using \( e^{-0.2} \approx 0.8187 \):

For \( x = 0 \): \[ P(X=0) = \frac{e^{-0.2} (0.2)^0}{0!} = 0.8187 \]
For \( x = 1 \): \[ P(X=1) = \frac{e^{-0.2} (0.2)^1}{1!} = 0.2 \times 0.8187 = 0.1637 \]
For \( x = 2 \): \[ P(X=2) = \frac{e^{-0.2} (0.2)^2}{2!} = \frac{0.04}{2} \times 0.8187 = 0.02 \times 0.8187 = 0.0164 \]

Summing them up:

\[ P(X \le 2) = 0.8187 + 0.1637 + 0.0164 = 0.9988 \]

Converting to percentage:

\[ 0.9988 \times 100 = 99.88\% \approx 99.9\% \]

Q15

In a Poisson distribution, if mean is \( 2 \), what is the variance?

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Answer: \( 2 \)

In a Poisson distribution, a key property is that the mean (\( \lambda \)) is equal to the variance.

\[ \text{Mean} = \lambda = 2 \] \[ \text{Variance} = \lambda \]

Therefore, Variance \( = 2 \).

Q16

It is given that \( 3\% \) defective electric bulb are manufactured by a company. Using Poisson distribution, find the probability of \( 100 \) bulbs will contain no defective bulbs. (Use \( e^{-3} = 0.05 \))

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Answer: \( 0.05 \)

Given:

Probability of defect \( p = 3\% = 0.03 \)
Number of trials (bulbs) \( n = 100 \)

In Poisson approximation to Binomial distribution, the mean \( \lambda \) is given by:

\[ \lambda = n \times p = 100 \times 0.03 = 3 \]

We need to find the probability of no defective bulbs, i.e., \( P(X=0) \).

\[ P(X=0) = \frac{e^{-\lambda} \lambda^0}{0!} \] \[ P(X=0) = \frac{e^{-3} \times 1}{1} = e^{-3} \]

Given \( e^{-3} = 0.05 \), so:

\[ P(X=0) = 0.05 \]

Q17

The mortality rate for a certain disease is \( 0.007 \). Using Poisson distribution, calculate the probability for \( 2 \) deaths in a group of \( 400 \) people.

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Answer: \( 0.24 \)

Given:

Probability of death \( p = 0.007 \)
Number of people \( n = 400 \)

Calculate mean \( \lambda \):

\[ \lambda = n \times p = 400 \times 0.007 = 2.8 \]

We need to find \( P(X=2) \):

\[ P(X=2) = \frac{e^{-2.8} (2.8)^2}{2!} \]

Using \( e^{-2.8} \approx 0.0608 \):

\[ P(X=2) = \frac{0.0608 \times 7.84}{2} \] \[ P(X=2) = \frac{0.47667}{2} \approx 0.238 \]

Q18

An ice-cream parlour receives a customer at an average rate of \( 4 \) per minute. If the number of customers received by the parlour follows a Poisson distribution, what is the approximate probability that \( 16 \) customers will be coming to the parlour in a particular \( 4 \)-minute period on a given day?

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Answer: \( 0.099 \)

Given:

Rate \( = 4 \) customers per minute.
Time period \( t = 4 \) minutes.

The mean number of customers for the \( 4 \)-minute period is:

\[ \lambda = \text{Rate} \times \text{Time} = 4 \times 4 = 16 \]

We need to find the probability of exactly \( 16 \) customers, \( P(X=16) \):

\[ P(X=16) = \frac{e^{-16} (16)^{16}}{16!} \]

Using Stirling's approximation or standard tables for Poisson distribution where \( \lambda = 16 \):

\[ P(X=16) \approx 0.0992 \]

Q19

Using Z-Table, Calculate:
a) \( P(Z < 1.20) \)
b) \( P(Z \le 1.20) \)
c) \( P(-0.5 \le Z \le 1.0) \)
d) \( P(-1.0 \le Z \le 1.0) \)

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Answer:

a) & b) \( P(Z < 1.20) \) and \( P(Z \le 1.20) \)

For continuous distributions like the Normal distribution, \( < \) and \( \le \) yield the same probability.

From standard Z-table, the area to the left of \( Z=1.20 \) is:

\[ P(Z < 1.20) = 0.8849 \]

c) \( P(-0.5 \le Z \le 1.0) \)

\[ P(-0.5 \le Z \le 1.0) = P(Z \le 1.0) - P(Z \le -0.5) \]

From Z-table:

\( P(Z \le 1.0) = 0.8413 \)
\( P(Z \le -0.5) = 1 - P(Z \le 0.5) = 1 - 0.6915 = 0.3085 \)

\[ \text{Probability} = 0.8413 - 0.3085 = 0.5328 \]

d) \( P(-1.0 \le Z \le 1.0) \)

\[ P(-1.0 \le Z \le 1.0) = P(Z \le 1.0) - P(Z \le -1.0) \]

From Z-table:

\( P(Z \le 1.0) = 0.8413 \)
\( P(Z \le -1.0) = 1 - 0.8413 = 0.1587 \)

\[ \text{Probability} = 0.8413 - 0.1587 = 0.6826 \]

Q20

A company conducted an IQ test for randomly select \( 50 \) employees. Volunteer A scored \( 74 \) out of the possible \( 120 \) points. If the average IQ test score was recorded as \( 62 \) and the standard deviation was \( 11 \). How well did volunteer A perform on the test compared to the other volunteers?

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Answer: Z-score = \( 1.09 \)

To compare performance, we calculate the Z-score using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

Given:

Score \( X = 74 \)
Mean \( \mu = 62 \)
Standard Deviation \( \sigma = 11 \)

Calculation:

\[ Z = \frac{74 - 62}{11} = \frac{12}{11} \approx 1.09 \]

Conclusion: Volunteer A's score is \( 1.09 \) standard deviations above the mean. This indicates they performed better than approximately \( 86\% \) of the participants.

Q21

An average ceiling fan manufactured by the Jagdeep Corporation lasts \( 300 \) days with a standard deviation of \( 50 \) days. Assuming that the ceiling fan's life is normally distributed, what is the probability that a ceiling fan will last at most \( 365 \) days?

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Answer: \( 0.9032 \)

Given:

Mean \( \mu = 300 \)
Standard Deviation \( \sigma = 50 \)
Value \( X = 365 \)

Calculate Z-score:

\[ Z = \frac{X - \mu}{\sigma} = \frac{365 - 300}{50} = \frac{65}{50} = 1.3 \]

We need to find \( P(X \le 365) \), which is equivalent to \( P(Z \le 1.3) \).

From the standard Z-table:

\[ P(Z \le 1.3) = 0.9032 \]

Q22

In a survey of daily travel time (in minutes) of students to reach their school was recorded as follows:
\( 26, 33, 65, 28, 34, 55, 25, 44, 50, 36, 26, 37, 43, 62, 35, 38, 45, 32, 28, 34 \)
If the mean travelling time is \( 38.8 \) minutes and the standard deviation is \( 11.4 \) minutes. Convert the travel time of each student into a Z-Score.

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Answer:

Formula: \( Z = \frac{X - 38.8}{11.4} \)

Calculations:

\( 26 \to Z = \frac{26-38.8}{11.4} = -1.12 \)
\( 33 \to Z = \frac{33-38.8}{11.4} = -0.51 \)
\( 65 \to Z = \frac{65-38.8}{11.4} = 2.30 \)
\( 28 \to Z = \frac{28-38.8}{11.4} = -0.95 \)
\( 34 \to Z = \frac{34-38.8}{11.4} = -0.42 \)
\( 55 \to Z = \frac{55-38.8}{11.4} = 1.42 \)
\( 25 \to Z = \frac{25-38.8}{11.4} = -1.21 \)
\( 44 \to Z = \frac{44-38.8}{11.4} = 0.46 \)
\( 50 \to Z = \frac{50-38.8}{11.4} = 0.98 \)
\( 36 \to Z = \frac{36-38.8}{11.4} = -0.25 \)
\( 37 \to Z = \frac{37-38.8}{11.4} = -0.16 \)
\( 43 \to Z = \frac{43-38.8}{11.4} = 0.37 \)
\( 62 \to Z = \frac{62-38.8}{11.4} = 2.04 \)
\( 35 \to Z = \frac{35-38.8}{11.4} = -0.33 \)
\( 38 \to Z = \frac{38-38.8}{11.4} = -0.07 \)
\( 45 \to Z = \frac{45-38.8}{11.4} = 0.54 \)
\( 32 \to Z = \frac{32-38.8}{11.4} = -0.60 \)

Q23

In an examination, \( 2000 \) students appeared and the mean of the normal distribution of marks is \( 30 \) with standard deviation as \( 6.25 \). Find out how many students are expected to score:
i. between \( 20 \) and \( 40 \) marks.
ii. less than \( 25 \) marks.

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Answer: i. \( 1781 \) students, ii. \( 424 \) students

Given: \( \mu = 30, \sigma = 6.25, N = 2000 \)

i. Between \( 20 \) and \( 40 \):

Calculate Z-scores:

\[ Z_1 = \frac{20 - 30}{6.25} = -1.6 \] \[ Z_2 = \frac{40 - 30}{6.25} = 1.6 \]

Probability \( P(-1.6 \le Z \le 1.6) \):

\[ P(Z \le 1.6) - P(Z \le -1.6) = 0.9452 - 0.0548 = 0.8904 \]

Number of students \( = 0.8904 \times 2000 = 1780.8 \approx 1781 \).

ii. Less than \( 25 \):

\[ Z = \frac{25 - 30}{6.25} = -0.8 \]

Probability \( P(Z < -0.8) \):

\[ P(Z < -0.8) = 0.2119 \]

Number of students \( = 0.2119 \times 2000 = 423.8 \approx 424 \).

Q24

In a normal distribution, \( 31\% \) of the articles are under \( 45 \) and \( 8\% \) are over \( 64 \). Calculate the mean and standard deviation of the distribution.

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Answer: Mean \( \approx 50 \), SD \( \approx 10 \)

Given:

\( P(X < 45) = 0.31 \)
\( P(X > 64) = 0.08 \Rightarrow P(X < 64) = 0.92 \)

Finding Z-scores from table:

For area \( 0.31 \), \( Z_1 \approx -0.496 \) (since \( 31\% \) is to the left).
For area \( 0.92 \), \( Z_2 \approx 1.405 \).

Forming equations:

\[ \text{(i) } 45 = \mu + (-0.496)\sigma \] \[ \text{(ii) } 64 = \mu + (1.405)\sigma \]

Subtract (i) from (ii):

\[ 19 = 1.901\sigma \Rightarrow \sigma \approx 10 \]

Substitute \( \sigma = 10 \) into (i):

\[ 45 = \mu - 4.96 \Rightarrow \mu \approx 49.96 \approx 50 \]