**PLEASE NOTE** - I am using Excel and it is CRUCIAL that I know the excel funct
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Question
**PLEASE NOTE** - I am using Excel and it is CRUCIAL that I know the excel functions to discover the answers. PLEASE HELP ME UNDERSTAND THIS BY LISTING THEM!!! Finding the answer isn't as hard for me as being able to input the functions via excel. THANK YOU SO MUCH!!!!
Trading volume on the New York Stock Exchange is heaviest during the first half hour (early morning) and last half hour (late afternoon) of the trading day. The early morning trading volumes (millions of shares) for 16 days in January and February are shown below. The probability distribution of trading volume is approximately normal. Label all answers appropriately.
a. Compute the mean and standard deviation to use as estimates of the population mean and standard deviation.
b. What is the probability that, on a randomly selected day, the early morning trading volume will be less than 180 million shares?
c. What is the probability that, on a randomly selected day, the early morning trading volume will exceed 200 million shares?
d. What is the probability that, on a randomly selected day, the early morning trading volume will be between 180 and 200 million shares?
d. How many shares would have to be traded for the early morning trading volume on a particular day to be among the busiest 5% of days?
**PLEASE NOTE** - I am using Excel and it is CRUCIAL that I know the excel functions to discover the answers. PLEASE HELP ME UNDERSTAND THIS BY LISTING THEM!!! Finding the answer isn't as hard for me as being able to input the functions via excel. THANK YOU SO MUCH!!!!
Volume 179 208 217 200 184 160 245 201 209 210 177 212 210 194 180 165Explanation / Answer
Answers:
a. Compute the mean and standard deviation to use as estimates of the population mean and standard deviation.
Mean = 196.9375 approximated to 197
Standard deviation = 21.81886 approximated to 22.
Take these values as Population mean and Population Standard deviation.
[use command =AVERAGE(select all the cells) to get average.
and use command STDEV(select all the cells) to get standard deviation.
b. What is the probability that, on a randomly selected day, the early morning trading volume will be less than 180 million shares?
Let X be a normal variate with mean µ and standard deviation .
Then is a Standard Normal Variate Z = (x-µ)/
In this problem X is the trading volume µ = 197 and = 22.
probability that, on a randomly selected day, the early morning trading volume will be less than 180 million shares
= P[x<180] = P [(x-µ)/ < (180-197)/22] = P[Z < -0.77]
= Area from (- infinity) to (-0.77)
= [area from 0 to infinity] – [area from 0 to -0.77] (Refer normal distribution table)
= 0.5 - 0.2794 = 0.2206
Therefore, the probability that, on a randomly selected day, the early morning trading volume will be less than 180 million shares is 0.2206.
c. What is the probability that, on a randomly selected day, the early morning trading volume will exceed 200 million shares?
Similar to the question subquestion b, but x = 200.
probability that, on a randomly selected day, the early morning trading volume will exceed 200 million shares
= P[x>200] = P [(x-µ)/ > (200-197)/22] = P[Z > 0.14]
= Area from 0.14 to infinity
= Area from to to infinity - Area from zert to 0.14
= 0.5 - 0.0557 = 0.4443
Therefore, probability that, on a randomly selected day, the early morning trading volume will exceed 200 million shares is 0.4443.
d. What is the probability that, on a randomly selected day, the early morning trading volume will be between 180 and 200 million shares?
probability that, on a randomly selected day, the early morning trading volume will be between 180 and 200 million shares =
P [ 180 < X < 200] = P [ -.77 < Z < 0,14 ] (values taken from subquestion b and c)
= Area from zero to 0.77 + Area from zero to 0.14 (Since normal distribution is symmetrical)
= 0.2794 + 0.0557 = 0.3351.
Therefore, probability that, on a randomly selected day, the early morning trading volume will be between 180 and 200 million shares is 0.3351.
d. How many shares would have to be traded for the early morning trading volume on a particular day to be among the busiest 5% of days?
(Its fifth question - the extra one!!!)
Here, we need to find the value of X for the given probability.
5% (0.05) of the busiest days signifies the higher trading volume.
P [(x-µ)/ = (X-197)/22] = P [Z = (X-197)/22 ] = 0.05
We have to find z value in the normal distribution table whcih accounts for probability of 0.45 (0.5 -0.05) which will be 1.645.
(X - 197)/ 22 = 1.645
Which results in the value of x to be 233 (Approximately)
Therefore, the number of shares would have to be traded for the early morning trading volume on a particular day to be among the busiest 5% of days are 233 million shares.
Hope this will help.
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