X is a normally distributed random variable with mean 46 and standard deviation 22.

Mathematics

1 answer:

sesenic [268]2 years ago

4 0

Answer:

B. 0.95

Step-by-step explanation:

The probability that the value of the random variable is in the given range is the area under the normal probability distribution curve over that range. The value of this area is conveniently provided by any number of calculators, apps, or spreadsheets. It can also be determined using the "empirical rule" that tells you the probability over certain normalized ranges.

<h3>normalized range</h3>

A standard normal probability density function (PDF) has a mean of 0 and a standard deviation of 1. The range in this problem can be normalized to some number of standard deviations above or below the mean. The mean is given as 46, and the standard deviation is given as 22.

In this problem, the <em>lower limit</em> of the range is ...

z = (2 -46)/22 = -2

standard deviations from the mean.

Similarly, the <em>upper limit</em> is ...

z = (90 -46)/22 = +2

standard deviations from the mean.

<h3>empirical rule</h3>

The empirical rule tells us that 95% of the area of the PDF lies between -2 and +2 standard deviations from the mean.

P(2 < X < 90) ≈ 0.95

_____

<em>Additional comment</em>

The empirical rule for other ranges is ...

± 1 standard deviation: 68%

±2 standard deviations: 95%

±3 standard deviations: 99.7%

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A study of the impact of executive networking on firm performance measured firm performance as annual return on equity (ROE), r

vova2212 [387]

Answer:

The value of ROE that will be exceeded by 78% of the firms is -1.77%.

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:

The mean ROE for the firms studied was 14.93% and the standard deviation was 21.74%. This means that $\mu = 0.1493, \sigma = 0.2174$