Question

Suppose that the probability function in the table below reflects the possible lifetimes (in months after emergence) for fruit flies. Let the random variable X measure fruit fly lifetimes (in months).

Fruit fly lifetimes (in months)

x	1	2	3	4	5	6
p(x)	0.35	?	0.15	0.10	0.10	0.05

(a) What proportion of fruit flies die in their second month? P(x=2)

(b) What is the probability that a fruit fly lives more than four months?
(c) What is the mean lifetime for a fruit fly?
(d) What is the standard deviation of fruit fly lifetimes?

Answer 1

Solution:

Given that,

a ) P( X = 2 )

P (x ) = 1

1 = P ( x = 1 ) + P ( x = 2 ) +   P ( x = 3 ) + P ( x = 4 ) + P ( x = 5 ) + P ( x = 6 )

1 = 0.35 +   P ( x = 2 ) + 0.15 + 0.10 + 0.10 +0.05

P ( x = 2 ) = 1 - 0.75 =0.25

P ( x = 2 ) = 0.25

b ) P( X > 4 )

P ( x = 5 ) + P ( x = 6 )

= 0.10 +0.05

= 0.15

P( X > 4 ) = 0.15

x	P(x)	x * P(x)	x2 * P(x)
1	0.35	0.35	0.35
2	0.25	0.5	1
3	0.15	0.45	1.35
4	0.1	0.4	1.6
5	0.1	0.5	2.5
6	0.05	0.3	1.8
	1	2.5	8.6

c ) Mean = = X * P(X) = 2.5

d) Standard deviation = =X ² * P(X) - ²  

=8.6 - 6.25

  =2.35

= 1.532