Answer to Question #129746 in Statistics and Probability for Hicks

Question

Melina recorded the number of passengers in each car of the first 100 cars that passed in front of her house on Sunday morning. The outcomes are shown in the table below:

Number of Passengers (x_i) 1 2 3 4 5
Number of Cars (f_i) 44 30 15 4 7


Find the mode(M_0) and median(x ̃) value?
Find the mean value(x ̅)?
Find the standard deviation (σ)? (Give your answer in two decimal places).
What is the probability a car to contain:
exactly 3 passengers?
at least 2 passengers?
at most 4 passengers?
exactly 6 passengers?
What kind of probability is the above problem? Justify your answer.

Answer 1

Solution:

The mode is most commonly occuring value.

Max(f_i)=f₁=44, so M_o=x₁=1

M_o=1

The mode value is equal 1.

The set of the ordered data has the form

{y₁,y₂,...,y₁₀₀}, where

y_i=1, i=1,2,..,f₁

f₁=44, so

y_i=1, i=1,2,..,44,

y_i=2, i=f₂+1,...,f₁+f₂

f₁+f₂=44+30, so

y_i=2, i=45,..,74,

and similarly

y_i=3, i=75,..,89

y_i=4, i=90,..,93

y_i=5, i=94,..,100

$\widetilde{x}$ =(y₅₀+y₅₁)/2=(2+2)/2=2

$\widetilde{x}=2$

The median value is equal 2.

$\overline{x}=(\sum$⁵_k=1 f_kx_k )/100

$\overline{x}=(44\cdot1+30\cdot2+15\cdot3+4\cdot4+7\cdot5)/100$

$\overline{x}=2$

The mean value is equal 2.

$\sigma^{2}$= ($\sum$⁵_k=1 f_k(x_k-$\overline{x}$ )²)/100

$\sigma^{2}$ =$(44\cdot(1-2)^{2}+30\cdot(2-2)^{2}+15\cdot(3-2)^{2}+4\cdot(4-2)^{2}+7\cdot(5-2)^{2})/100$

$\sigma^{2}=1.38$

$\sigma=\sqrt{1.38}$

$\sigma\approx1.17$

The standard deviation is approximately equal 1.17.

P(X=1)=f₁/100=44/100=0.44,

P(X=2)=f₂/100=30/100=0.3,

P(X=3)=f₃/100=15/100=0.15,

P(X=4)=f₄/100=4/100=0.04,

P(X=5)=f₅/100=7/100=0.07.

So, the probability that a car contains exactly 3 passengers is equal 0.15.

P(X$\geq$ 2)=P(X=2)+P(X=3)+P(X=4)+P(X=5)=0.3+0.15+0.04+0.07=0.56.

The probability that a car contains at least 2 passengers is equal 0.56.

P(X$\leq$4)=P(X=1)+P(X=2)+P(X=3)+P(X=4)=0.44+0.3+0.15+0.04=0.93.

The probability that a car contains at most 4 passengers is equal 0.93.

Since x₆=0, we have P(X=6)=0.

The probability that a car contains exactly 6 passengers is equal 0.

A random variable x_i takes a finite number value.

Therefore, in this problem we have dealt with discrete random variable.