In a random sample of 30 tires of the same type, it is found that the average life span is 36,200 miles with a standard deviation of 3800 miles.

Find the interval about the sample mean such that the probability is 0.77 that the mean number lies within the interval.
a. 35,402-36,998
b. 31,830-40,570
c. 35,436-36,964
d. 35,367-37,033

Is it C?

You are looking for the proportion that lie in the area μ ± .385 = μ ± 1.2 SD. (This information is available from a table in the back of your stats text called something like "areas under the normal distribution.")

However, for a distribution of means, you need the standard error rather than the standard deviation.

SE = SD/sq rt of n

So you are looking for μ ± 1.2SE

You should be able to calculate the values from this.

I hope this helps. Thanks for asking.

To solve this problem, we need to find the confidence interval around the sample mean.

First, let's calculate the standard error, which is the standard deviation divided by the square root of the sample size:

Standard Error = Standard Deviation / Square Root of Sample Size
= 3800 / sqrt(30)
≈ 693.11

Now, we can find the margin of error. Since we are looking for a 77% probability, the remaining probability outside the interval is (1 - 0.77) / 2 = 0.115. This is because the probability is evenly split on both sides of the interval.

To find the z-value corresponding to a 0.115 probability in the standard normal distribution, we can use a z-table or a calculator. From the table, the value is approximately 1.17.

Margin of Error = Z-value * Standard Error
= 1.17 * 693.11
≈ 810.44

Finally, we can calculate the confidence interval by subtracting and adding the margin of error from the sample mean:

Confidence Interval = Sample Mean ± Margin of Error
= 36200 ± 810.44
≈ (35389.56, 37010.44)

Comparing the given options, we can see that the correct answer is option c. (35,436-36,964) because it contains the confidence interval that we calculated.