Question

The lengths of a lawn mower part are approximately normally distributed with a given mean = 4 in. and standard deviation = 0.2 in. What percentage of the parts will have lengths between 3.8 in. and 4.2 in.? 34% 68% 95% 99.7%

Answer 1

Answer: EDGE 2020 Quiz

Explanation:

1. The average miles per gallon of a particular automobile model are approximately normally distributed with a given mean Mu = 43.8 miles per gallon and standard deviation Sigma = 5.1 miles per gallon. What percentage of the automobiles have an average miles per gallon between 38.7 miles per gallon and 48.9 miles per gallon?

A.) 68%

2. What is the mean of the normal distribution shown below?

C.) 4

3. The annual salaries of all employees at a financial company are normally distributed with a mean of $34,000 and a standard deviation of $4,000. What is the z-score of a company employee who makes an annual salary of $54,000?

D.) 5

4. The lengths of a lawn mower part are approximately normally distributed with a given mean Mu = 4 in. and standard deviation Sigma = 0.2 in. What percentage of the parts will have lengths between 3.8 in. and 4.2 in.?

B.) 68%

5. The mean of the temperatures in the chart is 24° with a standard deviation of 4°. Which temperature is within one standard deviation of the mean?

C.) 22°

6. Which statement must be true?

D.) Each distribution has the same mean and a different standard deviation.

7. The heights of a certain type of tree are approximately normally distributed with a mean height Mu = 5 ft and a standard deviation Sigma = 0.4 ft. Which statement must be true?

D.) A tree with a height of 6.2 ft is 3 standard deviations above the mean.

8. What is the difference of the means of the distributions?

A.) 15

9. )The weights of boxes of candies produced in a factory are normally distributed with a mean weight of 16 oz and a standard deviation of 1 oz. What is the weight of a box of candies with a z-score of 2?

B.) 18 oz

10. What is the mean of the normal distribution shown below?

B.) 0

Answer 2

68 %

Explanation:

Let's call X to the random variable ''lengths of a lawn mower part''.

X ~ N(given mean;standard deviation)

X ~ N(4 in;0.2 in)

To find the percentage,first we are going to turn this random variable X into a random variable Z. Z will be a N(0;1)

We do this by subtracting the given mean to the original variable and dividing by it deviation.

P (3.8 in < X < 4.2 in) =

P [(3.8 in - 4 in) / (0.2 in)] < [(X - 4 in)] / (0.2 in) < [(4.2 in - 4 in) / (0.2 in)]

P ( -1 < Z < 1 ) = P (Z < 1) - P (Z< -1)

Where we can find P (Z < 1) and P (Z< -1) in any table of a N(0;1)

P (Z < 1) - P (Z< -1) = 0.8413 - 0.1587 = 0.6826 = 68%