Question

1. The Ganubrabdrab Company has determined that the number of telephone calls it receives each hour during holidays follows a normal distribution with a mean of 80,000 and a standard deviation of 35,000. Suppose 60 holiday hours are chosen at random.

1. What is the sampling distribution for the 60 hours?
2. What is the probability that for these 60 hours the number of incoming calls will be greater than 91,970?
3. What is the probability that for these 60 hours the number of incoming calls will be less than 75,000?
4. Construct the normal curve for the number of telephone calls received?

Answer 1

Solution :

Given that,

mean = = 80000

standard deviation = = 35000

n = 60

= 80000

= / n = 35000 / 60 = 4518.4806

( a )

The sampling distribution of the mean = 80000

The standard deviation of the mean = 4518.4806

( b )

P( > 91970) = 1 - P( < 91970)

= 1 - P[( - ) / < ( 91970 - 80000) / 4518.4806 ]

= 1 - P(z < 2.649)

Using z table,    

= 1 - 0.9960

= 0.0040

Probability = 0.0040

( c )

P( < 75000 )

= P(( - ) / < ( 75000 - 80000) / 4518.4806 )

= P(z < -1.107 )

Using z table

= 0.1341

Probability = 0.1341

( d )

The normal curve for the number of telephone calls it receives each hour during holidays follows a normal distribution .