Respuesta :
Answer: .123
Step-by-step explanation: use random variable e-s which will follow a normal distribution with a mean of 70-60=10 and SD= sq root of 25+49. P(e-s<0)=.123
The probability that a randomly selected standard model tire will get more mileage than a randomly selected extended life tire is 0.123
What is the standard deviation?
It is defined as the measure of data disbursement, It gives an idea about how much is the data spread out.
We have the number of thousands of miles that the standard model tires last has a mean:
[tex]\rm \mu_S = 60[/tex]
And standard deviation:
[tex]\rm \sigma_s = 5[/tex]
The number of miles that the extended life tires last has a mean:
[tex]\rm \mu_E = 70[/tex]
And standard deviation:
[tex]\rm \sigma_E = 7[/tex]
As the mileages for both tires follow a normal distribution curve.
If we use the E-S random variables and these random variables follow the normal distribution.
Now:
[tex]\rm E =\mu_E- \mu_S[/tex]
E = 70 - 60 ⇒ 10 and
[tex]\rm S= \sqrt{\sigma_S ^2+\sigma_E^2}[/tex]
[tex]\rm S= \sqrt{5^2+7^2}[/tex]
S = √74 ⇒ 8.6023
Now from the normal distribution table the value of the probability:
=P(E - S<0)
= 0.123
Thus, the probability that a randomly selected standard model tire will get more mileage than a randomly selected extended life tire is 0.123
Learn more about the standard deviation here:
brainly.com/question/12402189
