Question

A company advertises two car tire models. The number of thousands of miles that the standard model tires last has a mean \mu_S = 60μ S ​ =60 and standard deviation \sigma_S = 5σ S ​ =5. The number of miles that the extended life tires last has a mean \mu_E = 70μ E ​ =70 and standard deviation \sigma_E = 7σ E ​ =7. If mileages for both tires follow a normal distribution, what is the probability that a randomly selected standard model tire will get more mileage than a randomly selected extended life tire?

Answer 1

Answer:

0.123

Step-by-step explanation:

Use the random variable E - S, which will follow a normal distribution with a mean of 70 - 60 = 10 and SD = \sqrt{5^2+7^}

P(E - S < 0) = 0.123

Answer 2

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 normally distributed data?

Normally distributed data is the distribution of probability which is symmetric about the mean.

The mean of the data is the average value of the given data. The standard deviation of the data is the half of the difference of the highest value and mean of the data set.

A company advertises two car tire models. The number of thousands of miles that the standard model tires last has a mean and standard deviation,

$\mu_S= 60$

$\sigma_S = 5$

The number of miles that the extended life tires last has a mean and standard deviation,

$\mu_E = 70$

$\sigma_E = 7$

Use the E-S random variable rule,

$E=\mu_E-\mu_s\\E=70-60\\E=10$

Similarly, the value of S,

$E=\sqrt{\sigma^2_E+\sigma^2_s}\\S=\sqrt{7^2+5^2}\\S=\sqrt{49+25}\\S=\sqrt{74}\\S=8.60$

For this E-S value, the value of probability from the table of normal distribution, we get as 0.123.

$P(E - S < 0)=0.123$

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 normally distributed data here;

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