Question

Let y denote the number of broken eggs in a randomly selected carton of one dozen eggs.

y 0 1 2 3 4

p(y) 0.59 0.25 0.10 0.04 0.02

(a)

Calculate and interpret μy.

μy =

(b)

In the long run, for what proportion of cartons is the number of broken eggs less than μy? (Round your answer to two decimal places).

Does this surprise you?

Yes

No

(c)

Explain why μy is not equal to

0 + 1 + 2 + 3 + 4

5

= 2.0.

This computation of the mean is incorrect because it does not take into account that the number of broken eggs are all equally likely.

This computation of the mean is incorrect because it does not take into account the probabilities with which the number of broken eggs need to be weighted.

This computation of the mean is incorrect because the value in the denominator should equal the maximum y value.

This computation of the mean is incorrect because it does not take into account the number of partially broken eggs.

This computation of the mean is incorrect because it includes zero in the numerator which should not be taken into account when calculating the mean.

Answer 1

Solution :

Let y denote the number of broken eggs in a randomly selected carton of one dozen eggs.

y 0 1 2 3 4

p(y) 0.59 0.25 0.10 0.04 0.02

(a)

Calculate and interpret μy.

μy = 0*0.59+1*0.25+2*0.10+3*0.04+4*0.02 = 0.65

μy = 0.65

(b)

In the long run, for what proportion of cartons is the number of broken eggs less than μy? (Round your answer to two decimal places).

P( x < μy ) = P ( x < 0.65 )

= P ( x = 0 )

=  0.59

Does this surprise you?

No

(c)

Explain why μy is not equal to

0 + 1 + 2 + 3 + 4

5

= 2.0.

This computation of the mean is incorrect because it does not take into account the probabilities with which the number of broken eggs need to be weighted.