Question

sume that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally distributed with a mean of $0.35 and a standard deviation of $0.33. Based on this information, what is the probability that a randomly selected stock will close up $0.75 or mor

Answer 1

Answer:

0.1131 = 11.31% probability that a randomly selected stock will close up $0.75 or more.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean of $0.35 and a standard deviation of $0.33.

This means that $\mu = 0.35, \sigma = 0.33$.

What is the probability that a randomly selected stock will close up $0.75 or more?

This is 1 subtracted by the p-value of Z when X = 0.75. So

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

$Z = \frac{0.75 - 0.35}{0.33}$

$Z = 1.21$

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

1 - 0.8869 = 0.1131

0.1131 = 11.31% probability that a randomly selected stock will close up $0.75 or more.