Question

. For a normal distribution, find the percentage of data that are a. Within 1 standard deviation of the mean.________ b. To the right of 1.5 standard deviations below the mean________ c. More than 0.5 standard deviations away from the mean

Answer 1

Answer:

a. The percentage of data within 1 standard deviation of the mean is 68.26%.

b. The percentage of data to the right of 1.5 standard deviations below the mean is of 93.32%.

c. The percentage of data more than 0.5 standard deviations away from the mean is of 61.7%.

Step-by-step explanation:

When the distribution is normal, we use 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.

a. Within 1 standard deviation of the mean

Between $Z = -1$ and $Z = 1$

This is the pvalue of Z = 1 subtracted by the pvalue of Z = -1.

Z = 1 has a pvalue of 0.8413

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

The percentage of data within 1 standard deviation of the mean is 68.26%.

b. To the right of 1.5 standard deviations below the mean

Greater than $Z = -1.5$, that is, 1 subtracted by the pvalue of Z = -1.5.

Z = -1.5 has a pvalue of 0.0668

1 - 0.0668 = 0.9332

The percentage of data to the right of 1.5 standard deviations below the mean is of 93.32%.

c. More than 0.5 standard deviations away from the mean

Below $Z = -0.5$ or above $Z = 0.5$. These percentages are the same, so we find one and multiply by 1.

Below Z = -0.5

This is the pvalue of Z = -0.5, which is 0.3085

0.3085*2 = 0.617

The percentage of data more than 0.5 standard deviations away from the mean is of 61.7%.