The distribution of heights of women aged 20 to 29 is approximately normal with a mean of 2.8 inches and standard deviation of 2.8 inches. The height (+/- 0.1 inch) of the middle 99.7% falls between a low of ?? inches and a high of ?? inches.
Please help on how I set this up to find the answer.
sorry. its 62.8 for the mean
.997 = mean ± Z * SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.003/2) and its Z score. Insert data into above equation and solve.
To find the answer, we can use the formula for finding the boundaries of a normal distribution based on the mean and standard deviation.
The middle 99.7% of a normal distribution is represented by three standard deviations on each side of the mean. In other words, we need to find the values that are 3 standard deviations below and above the mean.
First, we need to calculate the value of 3 standard deviations. Since the standard deviation is given as 2.8 inches, we can calculate 3 times the standard deviation as follows:
3 * 2.8 = 8.4 inches
Next, to find the low value, subtract the calculated value from the mean:
Low value = mean - 8.4 inches
Low value = 2.8 inches - 8.4 inches
Low value = -5.6 inches
Finally, to find the high value, add the calculated value to the mean:
High value = mean + 8.4 inches
High value = 2.8 inches + 8.4 inches
High value = 11.2 inches
Therefore, the height (+/- 0.1 inch) of the middle 99.7% falls between a low of -5.6 inches and a high of 11.2 inches.