Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 1.
Define Poisson distribution.
Solution:
Poisson distribution is a discrete frequency distribution which gives the probability of a number of independent events occurring in a fixed time. It is useful for characterizing events with very low probabilities of occurrence within some definite time or space.

Question 2.
Write any 2 examples for Poisson distribution.
Solution:
1. The number of alpha particles emitted by a radioactive substance in a fraction of a second.
2. Number of road accidents occurring at a particular interval of time per day.

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 3.
Write the conditions for which the Poisson distribution is a limiting case of the binomial distribution.
Solution:
Poisson distribution is a limiting case of binomial distribution under the following conditions:

  • the number of trials ‘n’ is indefinitely large i.e, → ∞
  • the probability of success ‘p’ in each trial is very small, i.e, p → 0
  • np = λ is finite. Thus p = \(\frac{\lambda}{n}\) and q = 1 – \(\frac{\lambda}{n}\), λ > 0

Question 4.
Derive the mean and variance of the Poisson distribution.
Solution:
Let X be a Poisson random variable with parameter λ. The p.m.f is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q4
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q4.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q4.2
Thus the mean and variance of Poisson distribution are both equal to λ.

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 5.
Mention the properties of Poisson distribution.
Solution:
Poisson distribution is the only distribution in which the mean and variance are equal.

Question 6.
The mortality rate for a certain disease is 7 in 1000. What is the probability for just 2 deaths on account of this disease in a group of 400? [Given e-2.8 = 0.06]
Solution:
Let X denote the number of deaths due to the disease
P(death) = \(\frac{7}{1000}\) = 0.007 ⇒ p = 0.007 and n = 400
The value of mean λ = np = (0.007) (400) = 2.8
Hence X follows a Poisson distribution with
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q6
So the probability of just 2 deaths on account of this disease in a group of 400 is 0.2352.

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 7.
It is given that 5% of the electric bulbs manufactured by a company are defective. Using Poisson distribution find the probability that a sample of 120 bulbs will contain no defective bulb.
Solution:
p(defective bulbs) = \(\frac { 5 }{100}\)
n = 120
The value of mean λ = np = 120 × \(\frac { 5 }{100}\)
λ = 6
Hence, x follows possion distribution with
P(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(x = 0) = \(\frac { e^{-6}(6)^0 }{0!}\) = e-6
= 0.0025

Question 8.
A car hiring firm has two cars. The demand for cars on each day is distributed as a Poisson variate, with a mean of 1.5. Calculate the proportion of days on which
(i) Neither car is used
(ii) Some demand is refused.
Solution:
Let X be the Poisson variable denoting the demand for the cars.
It is given that mean is 1.5 ⇒ λ = 1.5
(i) P (Neither car is used) = P (X = 0) = \(\frac{e^{-1.5}(1.5)^{0}}{0 !}=e^{-1.5}=0.2231\)
(ii) Some demand is refused when demand is more than 2 since the firm has only 2 cars. So we want P (X > 2)
Now P (X > 2) = 1 – P (X ≤ 2)
= 1 – [P(X = 2) + P(X = 1) + P(X = 0)]
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q8

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 9.
The average number of phone calls per minute into the switchboard of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be
(i) no phone at all
(ii) exactly 3 calls
(iii) at least 5 calls.
Solution:
Let X be the Poisson variable denoting the number of phone calls per minute.
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q9
P (X = 1) = \(\frac{e^{-2.5}(2.5)}{1 !}\) = (0.08208) (2.5) = 0.2052
Using the above values and P (X = 0) and P (X = 3) from the previous subdivisions in (A) we get,
P(X ≥ 5) = 1 – [0.1336 + 0.2138 + 0.2565 + 0.2052 + 0.08208]
= 1 – 0.89118
= 0.10882

Question 10.
The distribution of the number of road accidents per day in a city is Poisson with mean 4. Find the number of days out of 100 days when there will be
(i) no accident
(ii) at least 2 accidents and
(iii) at most 3 accidents.
Solution:
Let X be the Poisson variable denoting the number of accidents per day.
Given that mean is 4 (i.e,) λ = 4. The p.m.f is given by P(X = x) = \(\frac{e^{-4} 4^{x}}{x !}\)
(i) P (no accident) = P(X = 0) = e-4 = 0.0183
For 100 days we have 100 × 0.0183 = 1.83 ~ 2
Hence out of 100 days there will be no accident for 2 days.

(ii) P (atleast 2 accidents) = P (X ≥ 2)
= 1 – P (X < 2)
= 1 – [P(X = 1) + P(X = 0)]
= 1 – [e-4 (4) + e-4]
= 1 – (0.0183) (5)
= 1 – 0.0915
= 0.9085
For 100 days we have 100 × 0.9085 ~ 91
Hence out of 100 days there will be at least 2 accidents for 91 days.

(iii) P (atmost 3 accidents) = P (X ≤ 3)
= P (X = 0 ) + P (X = 1 ) + P (X = 2) + P (X = 3)
= \(e^{-4}\left[1+\frac{4}{1}+\frac{16}{2}+\frac{64}{6}\right]\)
= (0.0183) [23.6667]
= 0.4331
For 100 days we have 100 × 0.4331 ~ 43
Hence out of 100 days, there will be at most 3 accidents for 43 days.

Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 11.
Assuming that a fatal accident in a factory during the year is 1/1200, calculate the probability that in a factory employing 300 workers there will be at least two fatal accidents in a year, (given e-0.25 = 0.7788).
Solution:
Let X denote the number of accidents.
Given that probability of accidents ‘p’ is 1/1200 and n = 300
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q11
= 1 – [e-0.25 + e-0.25 (0.25)]
= 1 – e-0.25 (1.25)
= 1 – (0.7788) (1.25)
= 0.0265
Thus the probability that there will be atleast two fatal accidents in a year is 0.0265.

Question 12.
The average number of customers, who appear in a counter of a certain bank per minute is two. Find the probability that during a given minute
(i) No customer appears
(ii) three or more customers appear.
Solution:
Let X denote the number of customers.
Given λ = 2
(i) P (no customer) = P (X = 0)
= \(\frac{e^{-2}(2)^{0}}{0 !}\)
= e-2
= 0.1353
(ii) P (3 or more customers) = P (X ≥ 3)
Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q12
Thus during a given minute, the probability that three or more customers appear is 0.3235.