An examination is normally distributed with a mean score of 77 and a deviation of 6. find the percent of individuals scoring as indicated below:

between 71-83
between 83-65
above 89
less than 65
between 77 and 65

To find the percentages, we need to use the z-score formula in combination with the standard normal distribution table.

The z-score formula is: z = (x - μ) / σ
Where:
- x is the given value
- μ is the mean
- σ is the standard deviation

1. Between 71 and 83:
To find the percentage of individuals scoring between 71 and 83, we need to calculate the z-scores for both values and find the corresponding areas under the curve. Let's calculate the z-score for 71 and 83:

z-score for 71: z = (71 - 77) / 6 = -1
z-score for 83: z = (83 - 77) / 6 = 1

Now, we can use the standard normal distribution table (also known as the Z-table) to find the area between these z-scores. Looking up the z-scores in the table, we find:

The area to the left of z = -1 is 0.1587.
The area to the left of z = 1 is 0.8413.

To find the area between these two z-scores, we subtract the smaller area from the larger one:
0.8413 - 0.1587 = 0.6826

So, approximately 68.26% of individuals will score between 71 and 83 on the examination.

2. Between 83 and 65:
Using the same process as above, let's calculate the z-scores for 83 and 65:

z-score for 83: z = (83 - 77) / 6 = 1
z-score for 65: z = (65 - 77) / 6 = -2

Using the Z-table, we find:
The area to the left of z = 1 is 0.8413.
The area to the left of z = -2 is 0.0228.

To find the area between these two z-scores, we subtract the smaller area from the larger one:
0.8413 - 0.0228 = 0.8185

So, approximately 81.85% of individuals will score between 83 and 65 on the examination.

3. Above 89:
To find the percentage of individuals with scores above 89, we need to calculate the z-score for 89:

z-score for 89: z = (89 - 77) / 6 = 2

Using the Z-table, we find:
The area to the left of z = 2 is 0.9772.

However, we are interested in the area to the right of z = 2 (above 89). Since the total area under the curve is 1, we subtract the area to the left from 1:
1 - 0.9772 = 0.0228

So, approximately 2.28% of individuals will score above 89 on the examination.

4. Less than 65:
To find the percentage of individuals with scores less than 65, we need to calculate the z-score for 65:

z-score for 65: z = (65 - 77) / 6 = -2

Using the Z-table, we find:
The area to the left of z = -2 is 0.0228.

So, approximately 2.28% of individuals will score less than 65 on the examination.

5. Between 77 and 65:The percentage of individuals scoring between 77 and 65 is the same as the percentage of individuals scoring less than 77 minus the percentage of individuals scoring less than 65.

From the previous calculations:
Percentage of individuals scoring less than 77 is 0.5 or 50% (as it is the mean score).
Percentage of individuals scoring less than 65 is 0.0228 or 2.28%.

Therefore, the percentage of individuals scoring between 77 and 65 is approximately 50% - 2.28% = 47.72%.

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.