Question

The packaging process in a breakfast cereal company has been adjusted so that an average of μ = 13.0 oz of cereal is placed in each package. The standard deviation of the actual net weight is σ = 0.1 oz and the distribution of weights is known to follow the normal probability distribution.

1. what proportion of cereal packages contain more than 12.9 oz?

2. Suppose 25 cereal boxes are chosen at random, what is the mean weight of all possible sample means?

3.Suppose 25 cereal boxes are chosen at random, what is the standard deviation of all possible sample means X_bar (also known as standard error)? (round to 2 decimal digits, e.g. 0.11)

Answer 1

Solution :

Given that,

mean = = 13.0

standard deviation = = 0.1

1 ) P (x > 12.9 )

= 1 - P (x < 12.9 )

= 1 - P ( x -  / ) < ( 12.9 - 13.0 / 0.1)

= 1 - P ( z < -0.1 / 0.1 )

= 1 - P ( z < -1 )

Using z table

= 1 - 0.1587

= 0.8413

Probability = 0.8413

2 ) n =25

=13.0

3 ) ₌ / n = 0.1 25= 0.02