Solution:
The mode is most commonly occuring value.
Max(fi)=f1=44, so Mo=x1=1
Mo=1
The mode value is equal 1.
The set of the ordered data has the form
{y1,y2,...,y100}, where
yi=1, i=1,2,..,f1
f1=44, so
yi=1, i=1,2,..,44,
yi=2, i=f2+1,...,f1+f2
f1+f2=44+30, so
yi=2, i=45,..,74,
and similarly
yi=3, i=75,..,89
yi=4, i=90,..,93
yi=5, i=94,..,100
=(y50+y51)/2=(2+2)/2=2
The median value is equal 2.
5k=1 fkxk )/100
The mean value is equal 2.
= (5k=1 fk(xk- )2)/100
=
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 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(X4)=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 xi takes a finite number value.
Therefore, in this problem we have dealt with discrete random variable.
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