The number of trades (in thousands) completed daily by an online stock brokerage follows a normal distribution with a mean of 101.1 and a standard deviation of 26.5. On average, the brokerage receives $6.04 commission per trade.

For samples of size $ n = 15 $ days:

1. Determine the mean and standard deviation of the sampling distribution of the sample mean daily commissions received (in thousand dollars) accurate to 3 decimal places:
a) Mean = thousand dollars
b) Standard deviation = thousand dollars

2. Determine the following probabilities (as percentages) accurate to one decimal place:
a) What is the probability that the mean daily commissions received is more than $581,652? %
b) What is the probability that the mean daily commissions received is between $561,116 and $697,016? %

3. For the given sample size, what is the maximum average daily commissions receivable from the lowest 7.5% volume trading days? Round to the nearest thousand dollars.

A population of values has a normal distribution with $ \mu = 99.7 $ and $ \sigma = 2.9 $. You intend to draw a random sample of size $ n = 29 $.

1. Calculate the z-score, round it to two decimal places, then use the rounded z-score to determine the required probability accurate to four decimal places.
a) Find the probability that a single randomly selected value is less than 101.2.
\[ P(x < 101.2) = \]
b) Find the probability that a sample of size $ n = 29 $ is randomly selected with a mean less than 101.2.
\[ P(\bar{x} < 101.2) = \]

1. a) Mean = $6.04 thousand dollars
b) Standard deviation = $6.823 thousand dollars
2. a) Probability = 100%
b) Probability = 100%
3. Maximum average daily commissions receivable = $98,000

1. To determine the mean and standard deviation of the sampling distribution of the sample mean daily commissions received (in thousand dollars), we can use the following formulas:

a) Mean of the sampling distribution = Mean of the population = $6.04
b) Standard deviation of the sampling distribution = Standard deviation of the population / Square root of the sample size

= [tex]\frac{\$26.5}{\sqrt{15}} = \$6.823[/tex]

2. To find the probabilities, we need to use the z-score formula:

a) To find the probability that the mean daily commissions received is more than $581,652, we first need to find the z-score.

The z-score formula is given by:

[tex]z = \frac{x - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex],

where x is the value we want to find the probability for, μ is the mean of the population, σ is the standard deviation of the population, and n is the sample size.

z = [tex]\frac{\$581,652 - \$6.04}{\left(\frac{\$26.5}{\sqrt{15}}\right)}[/tex]

z = 21518.78

Using a standard normal distribution table or a calculator, we find that the probability is approximately 1.

b) To find the probability that the mean daily commissions received is between $561,116 and $697,016, we need to find the z-scores for both values.

z₁ = [tex]\frac{(\$561,116 - \$6.04)}{\left(\frac{\$26.5}{\sqrt{15}}\right)}[/tex]

z₁ = 20618.03

z₂ = [tex]\frac{\$697,016 - \$6.04}{\left(\frac{\$26.5}{\sqrt{15}}\right)}[/tex]

z₂ = 25618.23

Using the standard normal distribution table or a calculator, we find that the probability is approximately 1.

3. To find the maximum average daily commissions receivable from the lowest 7.5% volume trading days, we need to find the z-score that corresponds to a cumulative probability of 0.075.

Using the z-score formula and rearranging it to solve for x, we have: [tex]x = z \cdot \left(\frac{\sigma}{\sqrt{n}}\right) + \mu[/tex]

First, we find the z-score using the standard normal distribution table or a calculator. For a cumulative probability of 0.075, the z-score is approximately -1.405.

Then, we plug in the values to find x:

x = [tex]-1.405 \cdot \left(\frac{$2.9}{\sqrt{29}}\right) + $99.7[/tex]

x = $98.324

Rounding to the nearest thousand dollars, the maximum average daily commissions receivable from the lowest 7.5% volume trading days is $98,000.

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ZaraArmstrong ZaraArmstrong · Answer 2 · 2024-07-27T10:28:01-04:00

a)The mean of the sampling distribution is equal to the population mean, so it is 6.04 thousand dollars.

b)Standard deviation [tex]= 26.5 / \sqrt(15) = 6.830[/tex] thousand dollars.

the probability that the mean daily commissions received is between [tex]\$561,116[/tex]and [tex]\$697,016[/tex] is approximately [tex]99.77\% - 0.01\% = 99.76\%.[/tex]

To solve the given questions, we'll use the properties of the normal distribution.

1. Sampling Distribution:

a) Mean of the sampling distribution of the sample mean daily commissions received:

The mean of the sampling distribution is equal to the population mean, so it is 6.04 thousand dollars.

b) Standard deviation of the sampling distribution of the sample mean daily commissions received:

The standard deviation of the sampling distribution is equal to the population standard deviation divided by the square root of the sample size:

Standard deviation[tex]= 26.5 / \sqrt(15) = 6.830[/tex] thousand dollars.

2. Probability Calculations:

a) Probability that the mean daily commissions received is more than $581,652:

We'll calculate the z-score using the formula[tex]\[ z = \frac{{(\$581,652 - \$604)}}{{\left(\frac{{6.830}}{{\sqrt{15}}}\right)}} \approx -3.161 \][/tex]

Using a standard normal distribution table, we find that the probability is approximately 0.0008, or 0.08%.

b) Probability that the mean daily commissions received is between $561,116 and $697,016:

We'll calculate the z-scores for both values and find the probabilities associated with those z-scores:

[tex]z_1 = ($561,116 - $604) / (6.830 / \sqrt(15)) = -3.907[/tex]

[tex]z2 = ($697,016 - $604) / (6.830 / \sqrt(15)) = 2.870[/tex]

Using the standard normal distribution table, we find the probability associated with z1 as approximately 0.0001, or 0.01%, and the probability associated with z2 as approximately 0.9977, or 99.77%.

Therefore, the probability that the mean daily commissions received is between $561,116 and $697,016 is approximately 99.77% - 0.01% = 99.76%.

3. Maximum average daily commissions receivable from the lowest 7.5% volume trading days:

We'll find the z-score corresponding to the lowest 7.5%:

Using a standard normal distribution table, we find the z-score corresponding to the lowest 7.5% as approximately -1.15.

Now we'll find the corresponding value in the original scale:

[tex]\[ z = \frac{{(x - \mu)}}{\sigma} \][/tex]

[tex]-1.15 = (x - 99.7) / 2.9[/tex]

[tex]x - 99.7 = -1.15 * 2.9[/tex]

[tex]x - 99.7 = -3.335[/tex]

[tex]x = 96.365[/tex]

Rounding to the nearest thousand dollars, the maximum average daily commissions receivable is approximately [tex]\$96,000.[/tex]

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