Question

the distribution of speed follows an approximately normal distribution with a mean of 80 mph and a standard deviation of 7.7 mph. show your work! 5. what percent of the runs will give the delorean a speed less than 68.45 mph? 6. what percent of the runs will give the delorean a speed greater than 85 mph? 7. what percent of the runs will give the delorean a speed between 70 and 95 mph?

Explanation:

Step1: Calculate the z - score for \(x = 68.45\)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 80\), \(\sigma=7.7\), and \(x = 68.45\).

$$ z=\frac{68.45 - 80}{7.7}=\frac{- 11.55}{7.7}=-1.5 $$

Step2: Find the probability using the standard normal table

We want to find \(P(X\lt68.45)\), which is equivalent to \(P(Z\lt - 1.5)\).
Looking up \(z=-1.5\) in the standard - normal table (the cumulative distribution function of the standard normal distribution \(\varPhi(z)\)), we get \(P(Z\lt - 1.5)=0.0668\)

Step3: Calculate the z - score for \(x = 85\)

Using the formula \(z=\frac{x-\mu}{\sigma}\), with \(\mu = 80\), \(\sigma = 7.7\), and \(x = 85\)

$$ z=\frac{85 - 80}{7.7}=\frac{5}{7.7}\approx0.65 $$

Step4: Find the probability using the standard normal table

We want to find \(P(X\gt85)\), which is equivalent to \(P(Z\gt0.65)\)
Since \(P(Z\gt z)=1 - P(Z\lt z)\), and from the standard - normal table \(P(Z\lt0.65)=0.7422\)

$$ P(Z\gt0.65)=1 - 0.7422 = 0.2578 $$

Step5: Calculate the z - scores for \(x = 70\) and \(x = 95\)

For \(x = 70\):

$$ z_1=\frac{70 - 80}{7.7}=\frac{-10}{7.7}\approx - 1.30 $$

For \(x = 95\):

$$ z_2=\frac{95 - 80}{7.7}=\frac{15}{7.7}\approx1.95 $$

Step6: Find the probability using the standard normal table

We want to find \(P(70\lt X\lt95)\), which is equivalent to \(P(-1.30\lt Z\lt1.95)\)
Since \(P(a\lt Z\lt b)=P(Z\lt b)-P(Z\lt a)\)
From the standard - normal table, \(P(Z\lt - 1.30)=0.0968\) and \(P(Z\lt1.95)=0.9744\)

$$ P(-1.30\lt Z\lt1.95)=0.9744 - 0.0968=0.8776 $$

Answer:

1. \(6.68\%\)
2. \(25.78\%\)
3. \(87.76\%\)