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12 May, 20:28

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%

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  1. 12 May, 22:01
    0
    68 %

    Step-by-step 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%
  2. 12 May, 22:06
    0
    Answer: EDGE 2020 Quiz

    Step-by-step 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
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