Question

the height of 3-year-old girls is approximately normally distributed. emmie is 32.0 inches tall and is at the 32nd percentile of the distribution. diana is 34.0 inches tall and is at the 62nd percentile of the distribution. which of the following is closest to the mean of the height distribution? (m) 32.50 inches (n) 32.79 inches (a) 33.00 inches (b) 36.53 inches (c) 33.21 inches

Explanation:

Step1: Recall properties of normal distribution

In a normal distribution, the mean is the center (50th percentile). We know percentiles and corresponding heights: 32nd percentile is 32.0 inches, 62nd percentile is 34.0 inches. Let the mean be \(\mu\) and standard deviation be \(\sigma\). Using the z - score formula \(z=\frac{x - \mu}{\sigma}\), where \(x\) is the value, \(\mu\) is the mean, \(\sigma\) is the standard deviation.

For Emmie: \(z_{32}=\frac{32-\mu}{\sigma}\), and from standard normal table, the z - score for 32nd percentile (0.32 cumulative probability) is approximately \(z_{32}\approx - 0.475\) (since \(P(Z < - 0.475)\approx0.32\)).

For Diana: \(z_{62}=\frac{34 - \mu}{\sigma}\), and the z - score for 62nd percentile (0.62 cumulative probability) is approximately \(z_{62}\approx0.305\) (since \(P(Z < 0.305)\approx0.62\)).

Step2: Set up two equations

We have two equations:

1. \(32-\mu=- 0.475\sigma\) (from Emmie's data)
2. \(34 - \mu = 0.305\sigma\) (from Diana's data)

Subtract the first equation from the second equation:
\((34 - \mu)-(32 - \mu)=0.305\sigma-(- 0.475\sigma)\)
Simplify left side: \(34 - \mu - 32+\mu = 2\)
Simplify right side: \(0.305\sigma + 0.475\sigma=0.78\sigma\)
So, \(2 = 0.78\sigma\), then \(\sigma=\frac{2}{0.78}\approx2.564\)

Step3: Solve for the mean \(\mu\)

Substitute \(\sigma\approx2.564\) into the first equation \(32-\mu=- 0.475\times2.564\)
Calculate \(-0.475\times2.564\approx - 1.218\)
Then \(32-\mu=- 1.218\)
Solve for \(\mu\): \(\mu=32 + 1.218=33.218\approx33.21\)

Answer:

C. 33.21 inches