The level of nitrogen oxides (NOX) in the exhaust after 50,000 miles or fewer of driving of cars of a particular model varies Normally with mean 0.06 g/mi and standard deviation 0.01 g/mi. A company has 25 cars of this model in its fleet. What is the level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01? (Hint: This requires a backward Normal calculation. Round your answer to three decimal places.)

The level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01 is L = 0.065.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem:

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean and standard deviation, which is also called standard error

In this problem, we have that:

What is the level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01?

This is X when Z has a pvalue of 1-0.01 = 0.99. So it is X when Z = 2.325.

By the Central Limit Theorem

The level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01 is L = 0.065.

2.56 is the width of the garden

step-by-step explanation:

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42