2. Suppose that people’s weights are normally distributed, with mean 175 pounds and a standard deviation of 6 pounds. Round to the nearest hundredth of a percent

a. What percent of the population would weigh between 165 and 170 pounds?
b. What percent of the population would you expect to weigh more than 182 pounds?
c. What percent of the population would you expect to weigh between 172 and 180 pounds?
d. What percent of the population would you expect to be heavier than 163 pounds?

are you supposed to be using a table?

Here it is graphically..
http://davidmlane.com/hyperstat/z_table.html

that is all my worksheet says, im supposed to figure out the percent... not sure how to use the normal curve and im getting all messed up i don't know how to figure the percent

To solve these questions, we need to use the properties of the normal distribution. We can use the Z-score formula to standardize the values and then use a Z-table to find the corresponding percentiles.

a. To find the percent of the population weighing between 165 and 170 pounds, we need to calculate the Z-scores for these weights.

Z1 = (165 - 175) / 6
Z2 = (170 - 175) / 6

Using the Z-table, we can find the area under the curve between these Z-scores. Subtracting the smaller area from the larger area will give us the percentage.

b. To find the percent of the population weighing more than 182 pounds, we need to calculate the Z-score for this weight.

Z = (182 - 175) / 6

Using the Z-table, we can find the area to the left of this Z-score. Subtracting this area from 1 will give us the percentage.

c. To find the percent of the population weighing between 172 and 180 pounds, we need to calculate the Z-scores for these weights.

Z1 = (172 - 175) / 6
Z2 = (180 - 175) / 6

Using the Z-table, we can find the area under the curve between these Z-scores. Subtracting the smaller area from the larger area will give us the percentage.

d. To find the percent of the population weighing heavier than 163 pounds, we need to calculate the Z-score for this weight.

Z = (163 - 175) / 6

Using the Z-table, we can find the area to the left of this Z-score. Subtracting this area from 1 will give us the percentage.

Please provide the values for a, b, c, and d, and I will calculate the percentages step-by-step.

To solve these problems, we need to use the properties of the normal distribution. The normal distribution is a common probability distribution that is often used to model real-world phenomena.

In this case, we have a normal distribution with a mean (μ) of 175 pounds and a standard deviation (σ) of 6 pounds. To find the percentage of the population in different weight ranges, we need to calculate the area under the normal curve.

a. To find the percentage of the population that weighs between 165 and 170 pounds, we need to find the area under the curve between these two values. We can do this by calculating the z-scores for 165 and 170 and then finding the corresponding area using a standard normal distribution table or a calculator.

The z-score is a measure of how many standard deviations an observation is from the mean. It is calculated using the formula:

z = (x - μ) / σ

where x is the value we want to find the z-score for, μ is the mean, and σ is the standard deviation.

For 165 pounds:
z = (165 - 175) / 6 ≈ -1.67

For 170 pounds:
z = (170 - 175) / 6 ≈ -0.83

Next, we can look up the area under the standard normal curve for these z-scores. Using a standard normal distribution table or a calculator, we find that the area to the left of -1.67 is approximately 0.0475 and the area to the left of -0.83 is approximately 0.2031.

To find the area between these two z-scores, we subtract the smaller area from the larger area:

Area = 0.2031 - 0.0475 ≈ 0.1556

So, approximately 15.56% of the population would weigh between 165 and 170 pounds.

b. To find the percentage of the population that would weigh more than 182 pounds, we need to find the area to the right of this value under the normal curve.

Calculating the z-score for 182 pounds:

z = (182 - 175) / 6 ≈ 1.17

Using a standard normal distribution table or a calculator, we look up the area to the left of 1.17, which is approximately 0.8781.

To find the area to the right of 1.17 (i.e., weighing more than 182 pounds), we subtract the left area from 1:

Area = 1 - 0.8781 ≈ 0.1219

So, approximately 12.19% of the population would weigh more than 182 pounds.

c. To find the percentage of the population that would weigh between 172 and 180 pounds, we can follow a similar process as in part a.

Calculating the z-score for 172 pounds:

z = (172 - 175) / 6 ≈ -0.50

Calculating the z-score for 180 pounds:

z = (180 - 175) / 6 ≈ 0.83

Using the standard normal distribution table or a calculator, we find that the area to the left of -0.50 is approximately 0.3085 and the area to the left of 0.83 is approximately 0.7967.

To find the area between these two z-scores, we subtract the smaller area from the larger area:

Area = 0.7967 - 0.3085 ≈ 0.4882

So, approximately 48.82% of the population would weigh between 172 and 180 pounds.

d. To find the percentage of the population that would be heavier than 163 pounds, we need to find the area to the right of this value under the normal curve.

Calculating the z-score for 163 pounds:

z = (163 - 175) / 6 ≈ -2.00

Using the standard normal distribution table or a calculator, we find the area to the left of -2.00 is approximately 0.0228.

To find the area to the right of -2.00 (i.e., weighing more than 163 pounds), we subtract the left area from 1:

Area = 1 - 0.0228 ≈ 0.9772

So, approximately 97.72% of the population would be heavier than 163 pounds.