Question

The average number of customers visiting the science center was 800 per day last year and the populations standard deviation is 250 customers per day.

1. In a span of a month, i.e. 30 days, write out the distribution of the sample mean

2. What is the probability that the sample mean is over 275 customers per day in a month?

3. What is the probability that the sample mean is less than 275 customers per day in a month?

4. A good month means the average number of customers is more than the average of 95% of the other month. Determine the criteria of a good month.

Answer 1

Solution :

Given that ,

mean = = 800

standard deviation = = 250

n = 30

(1)

sampling distribution of the sample mean

= 800

sampling distribution of the standard deviation

= / n = 250 / 30 = 45.64

(2)

P( >275 ) = 1 - P( <275 )

= 1 - P[( - ) / < (275-800) /250 ]

= 1 - P(z <-2.1 )

Using z table

= 1 - 0.0179

= 0.9821

probability= 0.9821

(3)

P( <275 ) = 1 - P( <275 )

= P[( - ) / < (275-800) /250 ]

= P(z <-2.1 )

Using z table

= 0.0179

probability= 0.0179

(4)

Using standard normal table,

P(Z > z) = 95%

= 1 - P(Z < z) = 0.95

= P(Z < z ) = 1 - 0.95

= P(Z < z ) = 0.05

= P(Z < -1.65 ) = 0.05  

z = -1.65   

Using z-score formula  

= z * +   

= -1.65*45.64+800

= 724.694

= 723