Mathematics, 10.03.2020 18:01 noorshular

For 300 trading days, the daily closing price of a stock (in $) is well modeled by a Normal model with mean $196.64 and a standard deviation of $7.17. According to this model, what is the probability that on a randomly selected day in this period the stock price closed as follows:(a) Above $210.55?(b) Below $203.38?(c) Between $181.87 and $210.55?(d) Which would be more unusual, a day on which the stock price closed above $206 or below $190?

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Answer from: FoxodyGirl7134

a) 2.62% probability that on a randomly selected day in this period the stock price closed above $210.55.

b) 82.64% probability that on a randomly selected day in this period the stock price closed below $203.38.

c) 95.41% probability that on a randomly selected day in this period the stock price closed between $181.87 and $210.55.

d) Closed above $206

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 196.64, \sigma = 7.17$

According to this model, what is the probability that on a randomly selected day in this period the stock price closed as follows:

(a) Above $210.55?

This is 1 subtracted by the pvalue of Z when X = 210.55. So

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

$Z = \frac{210.55 - 196.64}{7.17}$

Z = 1.94

Z = 1.94 has a pvalue of 0.9738

1 - 0.9738 = 0.0262

2.62% probability that on a randomly selected day in this period the stock price closed above $210.55.

(b) Below $203.38?

This is the pvalue of Z when X = 203.38. So

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

$Z = \frac{203.38 - 196.64}{7.17}$

Z = 0.94

Z = 0.94 has a pvalue of 0.8264

82.64% probability that on a randomly selected day in this period the stock price closed below $203.38.

This is the pvalue of Z when X = 210.55 subtracted by the pvalue of Z when X = 181.87. So

X = 210.55

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

$Z = \frac{210.55 - 196.64}{7.17}$

Z = 1.94

Z = 1.94 has a pvalue of 0.9738

X = 181.87

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

$Z = \frac{181.87 - 196.64}{7.17}$

Z = -2.06

Z = -2.06 has a pvalue of 0.0197

0.9738 - 0.0197 = 0.9541

95.41% probability that on a randomly selected day in this period the stock price closed between $181.87 and $210.55.

(d) Which would be more unusual, a day on which the stock price closed above $206 or below $190?

The one event with the lower probability

Above $206

1 subtracted by the pvalue of Z when X = 206

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

$Z = \frac{206 - 196.64}{7.17}$

Z = 1.31

Z = 1.31 has a pvalue of 0.9049

1 - 0.9049 = 0.0951 = 9.51% probability of closing above $206.

Below $190

pvalue of Z when X = 190. So

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

$Z = \frac{190 - 196.64}{7.17}$

Z = -0.93

Z = -0.93 has a pvalue of 0.1762

17.62% probability of closing below $190.

The probability of closing above $206 is lower, so this is more unusual.

Answer from: Quest

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Answer from: Quest

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