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 80mph and a standard deviation of 7.7mph. What percentage of the runs will give the Delorean a speed greater than 87.7 mph? *

15.866%

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 87.7mph

μ is the population mean = 80 mph

σ is the population standard deviation = 7.7 mph

z = 87.7 - 80/7.7

z = 1

Probability value from Z-Table:

P(x<87.7) = 0.84134

P(x>87.7) = 1 - P(x<87.7)

= 1 - 0.84134

= 0.15866

Converting to percentage

= 0.15866 × 100

= 15.866%

The percentage of the runs that will give the Delorean a speed greater than 87.7 mph is 15.866%

i used to know how to do this in 7th but i forget easily

step-by-step explanation: