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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)

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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%

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