Respuesta :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 196.64, \sigma = 7.17[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{210.55 - 196.64}{7.17}[/tex]
[tex]Z = 1.94[/tex]
[tex]Z = 1.94[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{203.38 - 196.64}{7.17}[/tex]
[tex]Z = 0.94[/tex]
[tex]Z = 0.94[/tex] 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.
(c) Between $181.87 and $210.55?
This is the pvalue of Z when X = 210.55 subtracted by the pvalue of Z when X = 181.87. So
X = 210.55
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{210.55 - 196.64}{7.17}[/tex]
[tex]Z = 1.94[/tex]
[tex]Z = 1.94[/tex] has a pvalue of 0.9738
X = 181.87
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{181.87 - 196.64}{7.17}[/tex]
[tex]Z = -2.06[/tex]
[tex]Z = -2.06[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{206 - 196.64}{7.17}[/tex]
[tex]Z = 1.31[/tex]
[tex]Z = 1.31[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{190 - 196.64}{7.17}[/tex]
[tex]Z = -0.93[/tex]
[tex]Z = -0.93[/tex] 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.
