Question

a researcher studying frogs is investigating the distance that a certain species of frog can jump. the jump lengths appear to be approximately normally distributed with a mean of 90 inches and a standard deviation of 12 inches. directions: make a sketch of the \empirical rule\ for this setting. a) what is the probablity of observing a random frog jump is less than 78 inches? b) what jump lengths represent the middle 95% of frog jumps? between and c) what is the probability of observing a random frog jump that is longer than 114 inches?

Explanation:

Step1: Recall the empirical rule

The empirical rule for a normal distribution states that:

• Approximately \(68\%\) of the data lies within \(\mu\pm\sigma\)
• Approximately \(95\%\) of the data lies within \(\mu\pm2\sigma\)
• Approximately \(99.7\%\) of the data lies within \(\mu\pm3\sigma\)

Here, \(\mu = 90\) (mean) and \(\sigma=12\) (standard deviation)

Step2: Solve part (a)

First, find the \(z -\)score for \(x = 78\). The \(z-\)score formula is \(z=\frac{x-\mu}{\sigma}\)

$$ z=\frac{78 - 90}{12}=\frac{- 12}{12}=-1 $$

By the empirical rule, the area to the left of \(z=-1\) is \(\frac{100\% - 68\%}{2}+68\%/2=16\%\) or \(0.16\)

Step3: Solve part (b)

For the middle \(95\%\) of the data (by the empirical rule \(\mu\pm2\sigma\))
Lower bound: \(x_1=\mu - 2\sigma=90-2\times12=90 - 24 = 66\)
Upper bound: \(x_2=\mu + 2\sigma=90+2\times12=90 + 24=114\)

Step4: Solve part (c)

Find the \(z -\)score for \(x = 114\). Using \(z=\frac{x-\mu}{\sigma}\)

$$ z=\frac{114 - 90}{12}=\frac{24}{12}=2 $$

By the empirical rule, the area to the right of \(z = 2\) is \(\frac{100\% - 95\%}{2}=2.5\%\) or \(0.025\)

Answer:

a) \(0.16\)
b) \(66\) and \(114\)
c) \(0.025\)