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ankoles [38]
3 years ago
11

sume that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally dis

tributed 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
Mathematics
1 answer:
Darina [25.2K]3 years ago
4 0

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.

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