Question

A random variable X can assume only the values of 2 and 5. If its mean is 4 and 5. Find the probabilities of these points​

Answer 1

Answer:

$hi$

Step-by-step explanation:

Thus the value of k is

15

1

X=x

i

P(X=x

i

) XP X

2

P

1 k k k

2 2k 4k 8k

3 3k 9k 27k

4 4k 16k 64k

5 5k 25k 125k

∑XP=55k=55×

15

1

=3.66 ∑X

2

P=225k225×

15

1

=15

Mean of the given data (

x

ˉ

)=∑XP=3.66

Variance of the given data =∑X

2

P−

x

ˉ

2

=15−(3.66)

2

=15−13.4=1.6

Hence the mean and variance of the given data is 3.66 and 1.6 respectively.