The mean of a normal probability distribution is 100 and the standard

deviation is 8. About what percent of the observations lie between 84 and 116?

Use z-scores.

Formula:
z = (x - mean)/sd

You will need to calculate two z-scores:
z = (84 - 100)/8
z = (116 - 100)/8

You will see that one z-score is 2 standard deviations below the mean, and the second z-score is 2 standard deviations above the mean. Use a z-table to determine the probability between these two z-scores, then convert to a percent.

I hope this will help get you started.

Thanks for the guidance. Does it make sense that I would get 1?

1. z = (x - μ) / σ
z = (84-100)/8
z = -16/8
z = -2
and
z = (116-100)/8
z = 16/8
z = 2
area = .0228+.9772
Area = 1

To find the percentage of observations that lie between 84 and 116 in a normal probability distribution with a mean of 100 and a standard deviation of 8, you can use the Z-score formula and the Z-table.

The Z-score formula is given by:

Z = (X - μ) / σ

Where:
- Z is the Z-score
- X is the value (in this case, 84 and 116)
- μ is the mean (100 in this case)
- σ is the standard deviation (8 in this case)

First, let's find the Z-score for 84:

Z1 = (84 - 100) / 8
= -16 / 8
= -2

Next, let's find the Z-score for 116:

Z2 = (116 - 100) / 8
= 16 / 8
= 2

Once we have the Z-scores, we can now use the Z-table to find the probabilities associated with these Z-scores.

The Z-table provides the area under the standard normal curve for different Z-scores. Since the normal distribution is symmetric, you can find the area for a positive Z-score and subtract it from 1. This will give you the area between the two Z-scores.

Using the Z-table, the area to the left of Z = -2 is approximately 0.0228, and the area to the left of Z = 2 is approximately 0.9772.

To find the percentage between 84 and 116, subtract the area to the left of Z = -2 from the area to the left of Z = 2:

Area between Z = -2 and Z = 2 = 0.9772 - 0.0228 = 0.9544

Finally, convert this decimal to a percentage by multiplying by 100:

Percentage = 0.9544 * 100
= 95.44%

Approximately 95.44% of the observations lie between 84 and 116.