Question

assuming that the heights of female college students are normally distributed with mean 66 inches and standard deviation 2.9 inches, answer the following questions. (hint: use the figure below with mean μ and standard deviation σ.)
area under a normal curve
(a) what percentage of female college students are taller than 66 inches?

%
(b) what percentage of female college students are shorter than 66 inches?

%
(c) what percentage of female college students are between 63.1 inches and 68.9 inches?

%
(d) what percentage of female college students are between 60.2 and 71.8 inches?

Explanation:

Step1: Analyze the normal distribution symmetry

In a normal distribution, the mean divides the distribution into two equal halves. So, the percentage of data above the mean is \(50\%\) and below the mean is also \(50\%\).
For part (a), since \(66\) inches is the mean \(\mu\), the percentage of female college students taller than \(66\) inches is \(50\%\) (because the normal distribution is symmetric about the mean).
For part (b), the percentage of female college students shorter than \(66\) inches is \(50\%\) (due to the symmetry of the normal distribution about the mean).

Step2: Calculate the number of standard deviations

For part (c), calculate \(z\) - scores. The formula for \(z=\frac{x - \mu}{\sigma}\). For \(x = 63.1\), \(z_1=\frac{63.1 - 66}{2.9}=\frac{- 2.9}{2.9}=-1\). For \(x = 68.9\), \(z_2=\frac{68.9 - 66}{2.9}=\frac{2.9}{2.9}=1\). Looking at the empirical rule (the \(68 - 95 - 99.7\) rule), the percentage of data within \(z=-1\) and \(z = 1\) is \(68\%\).
For part (d), calculate \(z\) - scores. For \(x = 60.2\), \(z_1=\frac{60.2-66}{2.9}=\frac{- 5.8}{2.9}=-2\). For \(x = 71.8\), \(z_2=\frac{71.8 - 66}{2.9}=\frac{5.8}{2.9}=2\). According to the empirical rule, the percentage of data within \(z=-2\) and \(z = 2\) is \(95\%\).

Answer:

(a) \(50\%\)
(b) \(50\%\)
(c) \(68\%\)
(d) \(95\%\)