Question

will marty make it back to the future?
after accelerating for 20 seconds, a delorean sports car has a wide range of speeds that it can achieve, depending on traction. the distribution of speed follows an approximately normal distribution with a mean of 80 mph and a standard deviation of 7.7 mph.

1. label the appropriate values on the normal distribution
2. what percent of the runs will give the delorean a speed greater than 87.7 mph?
3. what percent of the runs will give the delorean a speed between 64.6 mph and 87.7 mph?
4. what percent of the runs will give the delorean a speed less than 68.45 mph?

Explanation:

Step1: Recall the empirical rule for normal distributions

The empirical rule states that for a normal distribution:

• Approximately \(68\%\) of the data lies within \(1\) standard deviation (\(\sigma\)) of the mean (\(\mu\)): \(\mu\pm\sigma\)
• Approximately \(95\%\) of the data lies within \(2\) standard deviations of the mean: \(\mu\pm2\sigma\)
• Approximately \(99.7\%\) of the data lies within \(3\) standard deviations of the mean: \(\mu\pm3\sigma\)

Given \(\mu = 80\) mph and \(\sigma=7.7\) mph

Step2: Solve for question 2

First, find how many standard deviations \(87.7\) is from the mean.
\(z=\frac{x - \mu}{\sigma}=\frac{87.7 - 80}{7.7}=\frac{7.7}{7.7} = 1\)
The total area under the normal curve is \(1\) (or \(100\%\)).
The area within \(\mu\pm1\sigma\) is \(68\%\). So the area outside \(\mu\pm1\sigma\) is \(100\% - 68\%=32\%\).
Since the normal distribution is symmetric, the area greater than \(\mu + 1\sigma\) is \(\frac{32\%}{2}=16\%\)

Step3: Solve for question 3

For \(x_1 = 64.6\), \(z_1=\frac{64.6 - 80}{7.7}=\frac{- 15.4}{7.7}=-2\)
For \(x_2 = 87.7\), \(z_2 = 1\) (calculated in step 2)
The area within \(\mu - 2\sigma\) and \(\mu+1\sigma\):
The area within \(\mu - 2\sigma\) and \(\mu+2\sigma\) is \(95\%\), so the area less than \(\mu - 2\sigma\) is \(\frac{100\% - 95\%}{2}=2.5\%\)
The area within \(\mu\pm1\sigma\) is \(68\%\), so the area greater than \(\mu + 1\sigma\) is \(16\%\) (from step 2)
The area we want is \(100\%-(2.5\% + 16\%)=81.5\%\)

Step4: Solve for question 4

For \(x = 68.45\), \(z=\frac{68.45 - 80}{7.7}=\frac{- 11.55}{7.7}=-1.5\)
We know that the area within \(\mu - 1\sigma\) and \(\mu+1\sigma\) is \(68\%\), so the area outside \(\mu\pm1\sigma\) is \(32\%\). The area within \(\mu - 1.5\sigma\) and \(\mu+1.5\sigma\):
The formula for the \(z -\)score cumulative distribution. Using the standard normal table (or the property that the area within \(\mu - 1.5\sigma\) and \(\mu+1.5\sigma\) is \(86.64\%\))
The area less than \(\mu - 1.5\sigma\) is \(\frac{100\% - 86.64\%}{2}=6.68\%\)

Answer:

1. \(16\%\)
2. \(81.5\%\)
3. \(6.68\%\)