Question

1.         Trading volume on the New York Stock Exchange has been growing in recent years. For the first two weeks of January 1998, the average daily volume was 646 million shares (Barron’s, January 1998). The probability distribution of daily volume is approximately normal with a standard deviation of about 100 million shares.

a.   What is the probability trading volume will be less than 400 million shares?

b.   What percentage of the time does the trading volume exceed 800 million shares?

3. If the exchange wants to issue a press release on the top 5% of trading days, what volume will trigger a release?

2. Suppose that the average client charge per hour for lawyers in the state of Iowa for out-of-court work is $125. Suppose further that a random telephone sample of 25 lawyers in Iowa is taken. If the population standard deviation is $50,

1. What is the probability of getting a sample mean equal or greater than $110 per hour?

2. What is the probability of getting a sample mean less than $100 per hour?

3. What is the probability of getting a sample mean of between $120 and $130 per hour?

Answer 1

1a) Let X be the trading volume

then

To find P( X < 400)

= P( z < -2.46)

= 0.0070 ( from z table)

Probability that trading volume will be less than 400 million shares is 0.0070

b) P( X > 800)

=P( z > 1.54 )

= 0.0618 ( from z table)

Percentage of trading volume exceed 800 million shares is 6.18%

c) We have to find c such that P( X > c ) = 0.05

From z table we find

P( z > 1.645 ) = 0.05

Thus we get

A volume of 810.5 million shares will trigger a release .

2a)Let X be the client charge per hour

with mean =$125 and standard deviation =$ 50

Let be the mean client charge per hour

The sampling distribution of sample mean follow Normal with mean = 125 ( population mean)

and standrad error =

that is

then

To find

= P( z > -1.5)

= 0.9332 ( from z table)

Probability that sample mean greater than equal to $110 is 0.9332

b)

= P( z <  -2.5)

= 0.0062  ( from z table)

Probability that sample mean less than $100 is 0.0062

c)

= P(-0.5 < z < 0.5)

= P( -0.5 < z < 0) +P( 0 < z < 0.5)

= 0.1915+0.1915 ( from z table)

= 0.3830  

Probability that sample mean is between $120 and $130 is 0.3830