Question

Carbon monoxide​ (CO) emissions for a certain kind of car vary with mean 3.704 ​g/mi and standard deviation 0.7 ​g/mi. A company has 70 of these cars in its fleet. Let y overbary represent the mean CO level for the​ company's fleet. ​a) What's the approximate model for the distribution of y overbary​? Explain. ​b) Estimate the probability that y overbary is between 3.8 and 3.9​g/mi. ​c) There is only a 11​% chance that the​ fleet's mean CO level is greater than what​ value?

Answer 1

a) Let Y = Carbon monoxide​ (CO) emissions for a certain kind of car

From the given information, the mean of Y is = = 3.704 g/mi, and the standard deviation of Y is = = 0.7 g/mi.

Let n = sample size = 70

Let = sample mean of the random variable Y

Since the sample size = n = 70 which is sufficiently large to apply the Central Limit Theorem(CLT).

By CLT, the sampling distribution of the sample mean() is approximately normal with following parameters.

Mean = = 3.704

Standard deviation of sample mean is

So from the above explanation the distribution of \bar y is approximately normal with mean = 3.704 and standard deviation = 0.083667
​
b) Estimate the probability that y overbary ()is between 3.8 and 3.9​g/mi.

Mathematically P(3.8 < < 3.9) = P( < 3.9)- P(< 3.8) = ...(1)

The formula of Z-score for the sample means() is as follows:

For = 3.9 , the z-score is

Using excel: The general command to find cumulative standard normal probability in excel is "=NORMSDIST(z)"

For z = 2.34262 ,

P(Z < 2.34262) = "=NORMSDIST(2.34262)" = 0.990358

Therefore P(< 3.9) = 0.990358 ...(2)

For =3.8 , the z-score is

For z = 1.147406 ,

P(Z < 1.147406) = "=NORMSDIST(1.147406)" = 0.874393

Therefore, P(< 3.8) = 0.874393 ...(3)

Plugging the values of (2) and (3) in equation (1), we get

P( 3.8< < ) = 0.990358 - 0.874393 = 0.116033

Therefore, P( 3.8<< 3.9) = 0.116033 (This is the final answer of part b)

c) There is only a 11​% chance that the​ fleet's mean CO level is greater than what​ value?

Here we want to find \bar y score = a such that

P( > a) = 0.11

Therefore, P(< a) = 1 - 0.11 = 0.89

Let's use excel:

the command to find the inverse normal random value is as follows:

"=NORMINV(p,mean,standard deviation)

So for probability p = 0.89, mean = 3.704, standard deviation = 0.083667, we have

a = "=NORMINV(0.89,3.704,0.083667)" = 3.80662 g/mi

So the required value = a = 3.80662 g/mi   (This is the final answer of part c)