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Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Compute the population mean margin of error for a 90% confidence interval when sigma is 7 and the sample size?

is 36

The area under a normal curve with mu = 35 and sigma = 7 is

f John gets an 90 on a physics test where the mean is 85 and the standard deviation is 3, where does he stand in relation to his classmates?

he is in the top 5%

he is in the top 10%

he is in the bottom 5%

he is in the bottom 1%

Find P(12 < x < 23) when mu = 19 and sigma = 6. Write your steps in probability notation

In a normal distribution with mu = 34 and sigma = 5 what number corresponds to z = -4?

Let’s assume you have taken 1000 samples of size 64 each from a normally distributed population. Calculate the standard deviation of the sample means if the population’s variance is 49.

Interpret a 93% confidence interval of (7.46, 12.84) for a population mean.

The area to the left of "z" is .5438. What z score corresponds to this area?

What is the critical z-value that corresponds to a confidence level of 86%?

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  • 1 decade ago
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    Compute the population mean margin of error for a 90% confidence interval when sigma is 7 and the sample size?

    is 36

    Margin of error = m

    Confidence level= 90%

    standard deviation = 7

    sample size = n = 36

    first, find z* or z star, at 90% confidence level, its 1.645

    m = 1.645*7/sqrt(36)

    = 1.645*7/6

    = 1.919

    m = 1.919

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    if John gets an 90 on a physics test where the mean is 85 and the standard deviation is 3, where does he stand in relation to his classmates?

    he is in the top 5%

    he is in the top 10%

    he is in the bottom 5%

    he is in the bottom 1%

    mean = 85

    standard deviation = 3

    x = 90

    find the z score, with the corresponding p value or the probability.

    z = ( 90 - 85) / 3 = 1.67

    1.67 is only the z score, you need to find the p on the z score table, and its 0.9525 or at 95.3th place from bottom.

    his above the 5% top in the class.

    Why ? because 1 - 0.9525 = 0.0475 or less than 5%

    Therefore, he is in the top 5%

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Find P(12 < x < 23) when mu = 19 and sigma = 6. Write your steps in probability notation

    the idea is using z score table, again.

    mean = 19

    standard deviation = 6

    at 12, z = (12 - 19)/ 6 = -1.17 or p = 0.1210

    at 23, z = (23-19)/6 = 0.67 or p = 0.7486

    in between, 0.7486 - 0.1210 = 0.6276 or 62.76%

    Answer: p = 0.6276 or 62.76%

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    In a normal distribution with mu = 34 and sigma = 5 what number corresponds to z = -4?

    mean = 34

    standard deviation = 5

    z = -4

    find x?

    (x - 34) / 5 = -4

    the idea is you just solve for x

    x - 34 = -20

    x = -20 + 34 = 14

    x = 14

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Let’s assume you have taken 1000 samples of size 64 each from a normally distributed population. Calculate the standard deviation of the sample means if the population’s variance is 49.

    population:

    variance = 49

    standard deviation = sqrt(s^2) = sqrt(49) = 7

    sample:

    standard deviation ?

    n = 1000

    standard deviation of sample=

    7/sqrt(1000) =0.2214

    s = 0.2214

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Interpret a 93% confidence interval of (7.46, 12.84) for a population mean.

    The idea is the true population mean will be 93% confident that be between 7.46 and 12.84 or we are 93% confident that the true mean of the population will be between 7.46 and 12.84.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    The area to the left of "z" is .5438. What z score corresponds to this area?

    Use the z score table to find the corresponding, the idea is the probability or the area to the left is 0.5438.

    p = 0.5438

    z from the table is 0.11

    z = 0.11

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    What is the critical z-value that corresponds to a confidence level of 86%?

    so I dont know the exact value, since my table doesnt expand into the 0.01% and I only have it between 1.645 and 1.960 which is between 80% and 90% confidence interval.

    *still confused? write back

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