Company Means and Standard Deviation While doing your calculations, Ameritrade h

Question

Company Means and Standard Deviation

While doing your calculations, Ameritrade has asked that you keep their most recent policy in mind. Ameritrade has instituted a policy in which firms with prices within ~68% of the normal distribution of scores are ignored, outside ~68% (but within ~95%) of the normal distribution of scores are monitored, and outside ~95% of the normal distribution of scores are restricted. This policy is based on the most recent fraudulent activities that took place at Firm X.

Provide the cutoff values you would use to ignore, monitor, or restrict trade for Firm A based on the recent policy implemented by Ameritrade. Use the numeric values in the following graphic as a guide.

HINT: You will need to use the means and SDs from the previous table and the 68-95-99 rule to create these cutoffs.

:

Given your calculations in the previous question, state the decision you would make (i.e., ignore, monitor, or restrict trade) if the stock price for Firm A changed to each of the three values presented below.

a) $1.00

b) $2.10

c) $0.85

Explanation / Answer

Part A.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 1.21
standard Deviation ( sd )= 0.8
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [1.21 ± 0.8]
= [ 1.21 - 0.8 , 1.21 + 0.8]
= [ 0.41 , 2.01 ]
About 95% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 2s.d)
So to the given normal distribution about 95% of the observations lie in between
= [1.21 ± 2 * 0.8]
= [ 1.21 - 2 * 0.8 , 1.21 + 2* 0.8]
= [ -0.39 , 2.81 ]
About 99% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 3s.d)
So to the given normal distribution about 99.7% of the observations lie in between
= [1.21 ± 3 * 0.8]
= [ 1.21 - 3 * 0.8 , 1.21 + 3* 0.8]
= [ -1.19 , 3.61 ]
part B.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 1.14
standard Deviation ( sd )= 0.73
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [1.14 ± 0.73]
= [ 1.14 - 0.73 , 1.14 + 0.73]
= [ 0.41 , 1.87 ]
About 95% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 2s.d)
So to the given normal distribution about 95% of the observations lie in between
= [1.14 ± 2 * 0.73]
= [ 1.14 - 2 * 0.73 , 1.14 + 2* 0.73]
= [ -0.32 , 2.6 ]
About 99% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 3s.d)
So to the given normal distribution about 99.7% of the observations lie in between
= [1.14 ± 3 * 0.73]
= [ 1.14 - 3 * 0.73 , 1.14 + 3* 0.73]
= [ -1.05 , 3.33 ]

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