Use the table to answer the question. Note: Round z-scores to the nearest hundredth and then find the required A values using the table.
The cholesterol levels of a group of young women at a university are normally distributed, with a mean of 181 and a standard deviation of 36. What percent of the young women have the following cholesterol levels? (Round your answers to one decimal place.)
(a) greater than 219
(b) between 184 and 222
(b) between 184 and 222
Well, well, well, we find ourselves waist-deep in the world of cholesterol levels, huh? Let's do some number juggling and find out what percent of these young women fall between 184 and 222.
First, we need to find the z-score for both 184 and 222. To do this, we use the formula:
z = (x - mean) / standard deviation
So for 184, we have:
z = (184 - 181) / 36
Now calculate that z-score and you'll get... *drumroll*... 0.08.
For 222, it's quite similar:
z = (222 - 181) / 36
When you crunch the numbers, you'll end up with a z-score of... *another drumroll*... 1.14.
Now, if you're hungry for some A-values, you can head to the table and look up the corresponding area under the normal curve for both of these z-scores. Once you do that, subtract the smaller area from the larger area, and *poof*, you'll have your answer!
To find the percentage of young women with cholesterol levels between 184 and 222, we need to find the area under the normal distribution curve between these two values.
Step 1: Calculate the z-scores for the given cholesterol levels using the formula:
z = (x - μ) / σ
where:
- x represents the cholesterol level,
- μ represents the mean, and
- σ represents the standard deviation.
For the given values:
- For 184:
z = (184 - 181) / 36
= 3 / 36
= 0.08 (rounded to the nearest hundredth)
- For 222:
z = (222 - 181) / 36
= 41 / 36
= 1.14 (rounded to the nearest hundredth)
Step 2: Refer to the z-table to find the area under the curve corresponding to each z-score.
Using the z-table, we find that:
- For a z-score of 0.08, the area is 0.5328.
- For a z-score of 1.14, the area is 0.8708.
Step 3: Calculate the required percentage by subtracting the smaller area from the larger area:
Percentage = (0.8708 - 0.5328) x 100
= 0.338 x 100
= 33.8%
Therefore, approximately 33.8% of the young women have cholesterol levels between 184 and 222.
To find the percentage of young women with cholesterol levels between 184 and 222, we need to find the area under the normal distribution curve between these two values.
First, we need to calculate the z-scores for these two cholesterol levels using the formula:
z = (x - μ) / σ
where x is the cholesterol level, μ is the mean, and σ is the standard deviation.
For the lower value, 184:
z = (184 - 181) / 36
z = 3 / 36
z = 0.08 (rounded to the nearest hundredth)
For the higher value, 222:
z = (222 - 181) / 36
z = 41 / 36
z = 1.14 (rounded to the nearest hundredth)
Next, we need to find the corresponding probabilities (areas) for these z-scores using a standard normal distribution table.
Looking up the z-score 0.08 on the table, we find that the area to the left of this z-score is approximately 0.5328.
Looking up the z-score 1.14 on the table, we find that the area to the left of this z-score is approximately 0.8729.
Now, to find the area between these two z-scores, we subtract the smaller area from the larger area:
0.8729 - 0.5328 = 0.3401
So, approximately 34.0% of the young women have cholesterol levels between 184 and 222.
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
To find your answers on the Z table, use this equation:
Z = (score-mean)/SD