Question

1. The length of my daughter’s tantrums is uniformly distributed between 5 and 10 minutes. Determine the following:

1. Proportion of tantrums that exceeds 7 minutes. 

2. Length exceeded by 90% of the tantrums.

Answer 1

Solution:

Let X denote the lenght of daughter's tantrums.

X ~ U(5,10)

So, f(x) = 1/5 , 5 X 10

= 0 , otherwise.

and F(x) = P(X x) =

(a) ->

Proportion of tantrums that exceeds 7 minutes

= P(X > 7)

= 1 - P(X 7)

= 1 - F(7)

= 1 - (7-5)/5

= 1 - 0.4

= 0.6

(b) ->

Let x be the lengh that is exceeded by 90% of the tantrums

Then,

P(X > x) = 0.90

or, 1 - P(X x) = 0.90

or P(X x) = 0.10

or,   = 0.10

or, x - 5 = 0.10*5

or, x = 5.5

Hence lenght of 5.5 minutes is exceeded by 90% of tantrums.