The distribution of heights of a population of adults is approximately normal with mean 66 inches and SD 2.5 inches.
1. Approximately what percent of the adults are over 6 feet tall?
2. Approximately what percent of the adults have heights that are within 1 inch of the average?
3. Approximately what percent of the adults are 70 inches tall, to the nearest inch?
4. Approximately what is the 90th percentile of the heights, in inches?
I REALLY NEED SOME HELP. Thank you in advance!
1. 6 feet = 72 inches
You need to find the z score
(number - mean) divided by SD
72 -66/2.5 = z score
Then use a standard z-table to find the area of under the curve to the right of that number. This number will be your %.
2. Find the z-score for 67 and 65. Find the area between those two scores using the z-table.
3. To the nearest inch. Are you looking for 69.5 to 70.4?
4. 90th percentile means that 90% of the values are below that number or 10% are above. Find the z-score from the table
set the z-score = (height - mean)/SD
Thank you Dr. Jane for giving a succinctly explanatory answer without doing all the work for HELPPPPP, although I do realize the deadline is approaching.
If you don't understand how to do a problem HELPPPPP, ask it on the discussion board and many will be happy to help you. And by the way, this was covered in lecture 5.
Hi guys, thanks for the above. I got all the equations right but my z score to percent was wrong. If I have a random answer z score. which I wont put due to honor code. of pretend 2.6, would my z score from the table be 2 on the left and .06 on the top (to find the intersection) or would I just use 2.6 and 0.0?? once I have that number - pretend its .9456 / would that be 94.56% or would it be 94.56-100 = 5.4%?? your help is soooooooooo appreciated.
No problem! I'm here to help. Let's break down each question step by step:
1. To find the percentage of adults over 6 feet tall, we need to convert 6 feet to inches. Since 1 foot is equal to 12 inches, 6 feet is equal to 6 x 12 = 72 inches.
Now, we need to calculate the z-score for 72 inches using the given mean and standard deviation. The z-score formula is: z = (x - μ) / σ, where x is the value we want to calculate the z-score for, μ is the mean, and σ is the standard deviation.
Plugging in the given values, we have: z = (72 - 66) / 2.5 = 2.4
Next, we need to find the proportion of the population above this z-score in a standard normal distribution table. Looking up the z-score of 2.4 in the table, we find the corresponding area to be approximately 0.9918.
Finally, we can convert this proportion to a percentage by multiplying it by 100. So, approximately 99.18% of the adults are over 6 feet tall.
2. To find the percentage of adults with heights within 1 inch of the average, we need to determine the range within which the heights fall.
The range can be calculated by adding and subtracting 1 inch from the mean height of 66 inches. This gives us a range of (66 - 1) to (66 + 1), which is 65 to 67 inches.
To find the proportion of the population within this range, we need to calculate the z-scores for the lower and upper limits of the range. Using the z-score formula:
Lower limit: z = (65 - 66) / 2.5 = -0.4
Upper limit: z = (67 - 66) / 2.5 = 0.4
Next, we need to find the area between these z-scores in the standard normal distribution table. Looking up the z-scores of -0.4 and 0.4 in the table, we find the corresponding areas of approximately 0.3446 and 0.6554, respectively.
The proportion within the range is the difference between the two areas: 0.6554 - 0.3446 = 0.3108.
Finally, we convert this proportion to a percentage by multiplying it by 100. So, approximately 31.08% of the adults have heights within 1 inch of the average.
3. To find the percentage of adults who are exactly 70 inches tall, we need to calculate the z-score for this height using the given mean and standard deviation.
Using the z-score formula: z = (70 - 66) / 2.5 = 1.6
Next, we look up the z-score of 1.6 in the standard normal distribution table to find the corresponding area. The table indicates that the area for a z-score of 1.6 is approximately 0.9452.
Finally, we convert this proportion to a percentage by multiplying it by 100. So, approximately 94.52% of the adults are 70 inches tall to the nearest inch.
4. To find the 90th percentile of the heights, we need to calculate the z-score that corresponds to the 90th percentile.
The z-score for a given percentile can be found using the standard normal distribution table. We want to find the z-score where the area to the left of the z-score is 0.90 or 90%.
Looking up the z-score for a cumulative area of 0.90, we find that it is approximately 1.28.
Now, we can use the z-score formula to calculate the height corresponding to this z-score:
z = (x - μ) / σ
1.28 = (x - 66) / 2.5
Rearranging the equation, we have:
1.28 * 2.5 = x - 66
3.2 = x - 66
x = 69.2
So, approximately the 90th percentile of adult heights is 69.2 inches.