Question

7. 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 13 days in January and February are shown here (Barron’s, January 23, 2006; February 13, 2006; and February 27, 2006). 214 163 265 194 180 202 198 212 201 174 171 211 211 The probability distribution of trading volume is approximately normal. 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 230 million shares?

Answer 1

In this problem, first we find the sample mean and the standard deviation, and then, we use the normal distribution. Then, we get that:

a) The mean is of 199.69 and the standard deviation is of 25.02.

b) 0.2148 = 21.48% probability that, on a randomly selected day, the early morning trading volume will be less than 180 million shares.

c) 0.1131 = 11.31% probability that, on a randomly selected day, the early morning trading volume will exceed 230 million shares.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

Item a:

The mean is the sum of all observations divided by the number of observations, thus:

$\mu = \frac{214 + 163 + 265 + 194 + 180 + 202 + 198 + 212 + 201 + 174 + 171 + 211 + 211}{13} = 199.69$

The standard deviation is given by the square root of the sum of the difference squared between each observation and the mean, divided by the number of values. Thus:

$\sigma = \sqrt{\frac{(214-199.69)^2 + (163-199.69)^2 + (265-199.69)^2 + (194-199.69)^2 + ... + (211-199.69)^2}{13}} = 25.02$

The mean is of 199.69 and the standard deviation is of 25.02.

Item b:

This probability is the p-value of Z when X = 180, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{180 - 199.69}{25.02}$

$Z = -0.79$

$Z = -0.79$ has a p-value of 0.2148.

0.2148 = 21.48% probability that, on a randomly selected day, the early morning trading volume will be less than 180 million shares.

Item c:

This probability is 1 subtracted by the p-value of Z when X = 230, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{230 - 199.69}{25.02}$

$Z = 1.21$

$Z = 1.21$ has a p-value of 0.8869.

1 - 0.8869 = 0.1131

0.1131 = 11.31% probability that, on a randomly selected day, the early morning trading volume will exceed 230 million shares.

A similar problem is given at brainly.com/question/24663213