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    1. Suppose you have drawn a simple random sample of 10 students from a college campus and recorded how many hours each student surfed the internet during the first week of February, 2010.

    Results:
    Student: 1 2 3 4 5 6 7 8 9 10
    Hours of Internet Surfing: 9 12 4 10 5 18 8 12 6 6

    For this sample, compute (Xi), the sample average (X-bar), (Xi - X-bar)2, and the sample standard deviation (s).

    2. Suppose the annual snowfall in a city is normally distributed with a mean of 80 inches and a standard deviation of 25 inches. Find the probability that in a given year, snowfall in the city would be between 60 and 100 inches (that is, within ± 20 of m = 80).

    3. Let X denote the amount of money an SU student spends on books in a year. Assume that population mean is $800, and the population standard deviation is $125. Suppose you have drawn a simple random sample of size 400 from the SU student
    population. Compute the probability that the sample mean (X) is between $790 and $810.

    4. Suppose 20% of SU students own Apple computers, that is, p = 0.2. You have drawn a simple random sample of size 400 from this population. Let p denote the proportion of the sample that own Apple computers. Compute the probability that for your sample of 400 students, p will fall between 0.18 to 0.22 (i.e., within ±.02 of p = .2)

    © BrainMass Inc. brainmass.com December 24, 2021, 9:33 pm ad1c9bdddf
    https://brainmass.com/statistics/sampling/statistics-probability-sample-means-392002

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    4. Suppose you have drawn a simple random sample of 10 students from a college campus and recorded how many hours each student surfed the internet during the first week
    of February, 2010.

    Results:

    Student 1 2 3 4 5 6 7 8 9 10
    Hours of Internet Surfing 9 12 4 10 5 18 8 12 6 6

    For this sample, compute ∑ Xi, the sample average (X-bar), ∑ (Xi - X-bar)2, and the sample standard deviation (s).

    Solution:

    ∑ Xi = (9+12+4+10+......+6+6) = 90

    (X-bar) = ∑ Xi /n = 90/10 = 9

    x_i x ̅ ( x_i-x ̅ ) ( x_i-x ̅ )2
    9 9 (9-9) = 0 (0)2 = 0
    12 9 (12-9) = 3 (3)2 = 9
    4 9 (4-9) = -5 (-5)2 = 25
    10 9 (10-9) = 1 (1)2 = 1
    5 9 (5-9)= -4 (-4)2 = 16
    18 9 (18-9)= 9 (9)2 = 81
    8 9 (8-9) = -1 (-1)2 = 1
    12 9 (12-9) = 3 (3)2 = 9
    6 9 (6-9) = -3 (-3)2 = 9
    6 9 (6-9) = -3 (-3)2 = 9

    ∑ (Xi - X-bar)2 = 0+9+25+1+.....+9+9 = 160

    Sample Standard Deviation =
    S= √({ (∑_(i=1)^n▒(x_i-x ̅ )^2 )/(n-1)} )=√(160/((10-1)))=4.2163

    5. Suppose the annual snowfall in a city is normally distributed with a mean of 80 inches
    and a standard deviation of 25 inches.

    Find the probability that in a given year, snowfall in the city would be between 60 and 100 inches (that is, within ± 20 of μ = 80).

    Solution:
    Mean,μ=80
    Standard Deviation,σ=25
    z-score=(x-μ)/σ

    For x=60,z-score=(x-μ)/σ=(60-80)/25=-0.8
    For x=100,z-score=(x-μ)/σ=(100-80)/25=0.8

    P(60<x<100)=P(-0.8<Z<0.8)=0.2881+0.2881=0.5762
    Probability (in terms of percentage)=57.62%

    6. Let X denote the amount of money an SU student spends on books in a year. Assume
    that population mean (μ) of X is $800, and the population standard deviation (σ) of X is
    $125. Suppose you have drawn a simple random sample of size 400 from the SU student
    population. Compute the probability that the sample mean (X) is between $790 and $810.

    Solution:

    Population Mean,μ=800
    Population Standard Deviation,σ=125

    Sample Size,n=400
    Sample Mean,μ ̅=Population Mean,μ=800
    Sample Standard Deviation,σ ̅=σ/√n =125/√400=6.25

    z-score=(x-μ ̅)/σ ̅
    For x=790,z-score=(790-800)/6.25=-1.6
    For x=810,z-score=(810-800)/6.25=1.6

    P(790<x<810)=P(-1.6<Z<1.6)=0.4452+0.4452=0.8904
    Probability (in terms of percentage)=89.04%

    7. Suppose 20% of SU students own Apple computers, that is, π = 0.2. You have drawn
    a simple random sample of size 400 from this population. Let p denote the proportion of
    the sample that own Apple computers. Compute the probability that for your sample of 400
    students, p will fall between .18 to .22 (i.e., within ±.02 of π = .2).

    Solution:

    p=0.2
    q=1-p=1-0.2=0.8
    Sample Size,n=400
    Mean,μ=np=400×0.2=80
    Standard Deviation,σ =√npq=√(400×0.2×0.8)=√64=8

    z-score=(x-μ)/σ

    For π=0.18,x=400×0.18=72,z-score=(x-μ)/σ=(72-80)/8=-1
    For π=0.22,x=400×0.22=88,z-score=(x-μ)/σ=(88-80)/8=+1

    P(0.18<π<0.22)=P(72<x<88)=P(-1<Z<1)
    =0.3413+0.3413=0.6826
    Probability (in terms of percentage)=68.26%

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 9:33 pm ad1c9bdddf>
    https://brainmass.com/statistics/sampling/statistics-probability-sample-means-392002

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