# Statistics

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 Â± 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 Â±.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 âˆ‘ Xi, the sample average (X-bar), âˆ‘ (Xi - X-bar)2, and the sample standard deviation (s).

Solution:

âˆ‘ Xi = (9+12+4+10+......+6+6) = 90

(X-bar) = âˆ‘ Xi /n = 90/10 = 9

x_i x Ì… ( x_i-x Ì… ) ( x_i-x Ì… )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

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

Sample Standard Deviation =

S= âˆš({ (âˆ‘_(i=1)^nâ–’(x_i-x Ì… )^2 )/(n-1)} )=âˆš(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 Â± 20 of Î¼ = 80).

Solution:

Mean,Î¼=80

Standard Deviation,Ïƒ=25

z-score=(x-Î¼)/Ïƒ

For x=60,z-score=(x-Î¼)/Ïƒ=(60-80)/25=-0.8

For x=100,z-score=(x-Î¼)/Ïƒ=(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 (Î¼) of X is $800, and the population standard deviation (Ïƒ) 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,Î¼=800

Population Standard Deviation,Ïƒ=125

Sample Size,n=400

Sample Mean,Î¼ Ì…=Population Mean,Î¼=800

Sample Standard Deviation,Ïƒ Ì…=Ïƒ/âˆšn =125/âˆš400=6.25

z-score=(x-Î¼ Ì…)/Ïƒ Ì…

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, Ï€ = 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 Â±.02 of Ï€ = .2).

Solution:

p=0.2

q=1-p=1-0.2=0.8

Sample Size,n=400

Mean,Î¼=np=400Ã—0.2=80

Standard Deviation,Ïƒ =âˆšnpq=âˆš(400Ã—0.2Ã—0.8)=âˆš64=8

z-score=(x-Î¼)/Ïƒ

For Ï€=0.18,x=400Ã—0.18=72,z-score=(x-Î¼)/Ïƒ=(72-80)/8=-1

For Ï€=0.22,x=400Ã—0.22=88,z-score=(x-Î¼)/Ïƒ=(88-80)/8=+1

P(0.18<Ï€<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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