Question

georgias k - 12 mathematics standards
a) what percent of women are taller than 70.5 inches?
b) between what heights do the middle 95% of women fall?
c) what percent of women are shorter than 63 inches?
d) a height of 68 inches corresponds to what percentile of adult female american heights?

1. this is an example of a probability histogram of a continuous random variable with an approximate mean of five ($\mu = 5$) and standard deviation of one ($\sigma = 1$). a normal curve with the same mean and standard deviation has been overlaid on the histogram.

a) what do you notice about these two graphs?

1. first, you must realize that normally distributed data may have any value for its mean and standard deviation. below are two graphs with three sets of normal curves.

Explanation:

1. Assume the heights of women follow a normal - distribution with mean \(\mu\) and standard deviation \(\sigma\). Commonly, for adult female American heights, \(\mu = 64.5\) inches and \(\sigma = 2.5\) inches. We will use the z - score formula \(z=\frac{x - \mu}{\sigma}\), where \(x\) is the value from the data set, \(\mu\) is the mean, and \(\sigma\) is the standard deviation, and the standard normal distribution table (z - table) to solve the problems:

• a) What percent of women are taller than 70.5 inches?
• Step 1: Calculate the z - score
• Given \(\mu = 64.5\), \(\sigma = 2.5\), and \(x = 70.5\). Using the z - score formula \(z=\frac{x-\mu}{\sigma}\), we have \(z=\frac{70.5 - 64.5}{2.5}=\frac{6}{2.5}=2.4\).
• Step 2: Find the area to the left of \(z = 2.4\) in the z - table
• Looking up \(z = 2.4\) in the standard normal distribution table, the area to the left \(P(Z\lt2.4)=0.9918\).
• Step 3: Find the area to the right of \(z = 2.4\)
• The area to the right \(P(Z > 2.4)=1 - P(Z\lt2.4)=1 - 0.9918 = 0.0082\). So the percentage of women taller than 70.5 inches is \(0.82\%\).
• b) Between what heights do the middle 95% of women fall?
• Step 1: Determine the z - scores for the boundaries
• For a 95% confidence interval in a normal distribution, the area in the two - tails is \(1 - 0.95=0.05\), so the area in each tail is \(\frac{0.05}{2}=0.025\). The z - score corresponding to an area of \(0.025\) to the left is \(z=- 1.96\), and the z - score corresponding to an area of \(0.975\) ( \(0.5+0.475\)) to the left is \(z = 1.96\).
• Step 2: Use the z - score formula to find the heights
• For \(z=-1.96\), we have \(-1.96=\frac{x_1 - 64.5}{2.5}\), then \(x_1=64.5-1.96\times2.5=64.5 - 4.9 = 59.6\) inches.
• For \(z = 1.96\), we have \(1.96=\frac{x_2 - 64.5}{2.5}\), then \(x_2=64.5 + 1.96\times2.5=64.5+4.9 = 69.4\) inches. So the middle 95% of women fall between 59.6 inches and 69.4 inches.
• c) What percent of women are shorter than 63 inches?
• Step 1: Calculate the z - score
• Using the z - score formula with \(x = 63\), \(\mu = 64.5\), and \(\sigma = 2.5\), we get \(z=\frac{63 - 64.5}{2.5}=\frac{-1.5}{2.5}=-0.6\).
• Step 2: Find the area to the left of \(z=-0.6\) in the z - table
• Looking up \(z=-0.6\) in the standard normal distribution table, \(P(Z\lt - 0.6)=0.2743\). So the percentage of women shorter than 63 inches is \(27.43\%\).
• d) A height of 68 inches corresponds to what percentile of adult female American heights?
• Step 1: Calculate the z - score
• Using the z - score formula with \(x = 68\), \(\mu = 64.5\), and \(\sigma = 2.5\), we have \(z=\frac{68 - 64.5}{2.5}=\frac{3.5}{2.5}=1.4\).
• Step 2: Find the area to the left of \(z = 1.4\) in the z - table
• Looking up \(z = 1.4\) in the standard normal distribution table, \(P(Z\lt1.4)=0.9192\). So a height of 68 inches corresponds to the 91.92nd percentile.

1. a) What do you notice about these two graphs?

• Brief Explanation: The probability histogram of the continuous random variable has a shape that is approximately bell - shaped, similar to the normal curve that is overlaid on it. The normal curve provides a smooth approximation of the distribution of the data represented by the histogram.
• Answer: The probability histogram and the normal curve have a similar bell - shaped form, and the normal curve is a smooth approximation of the histogram.

Answer:

a) \(0.82\%\)
b) 59.6 inches and 69.4 inches
c) \(27.43\%\)
d) 91.92nd percentile

1. a) The probability histogram and the normal curve have a similar bell - shaped form, and the normal curve is a smooth approximation of the histogram.