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#5 Sketch of Solutions
#1
X = the # of 50 students who graduate in 6 years. X is Binomial n = 50 and p = 0.5
Probability Density Function
Binomial with n = 50 and p = 0.5
x
22
P( X = x )
0.0788257
the probability that 22 students in the sample graduate in 6 years is 0.0788257
Cumulative Distribution Function
Binomial with n = 50 and p = 0.5
x
22
P( X ≤ x )
0.239944
the probability that 22 or fewer students in the sample graduate in 6 years is 0.239944
the probability that more than 22 students in the sample graduate in 6 years is 1 – 0.239944 = 0.760056
Cumulative Distribution Function
Binomial with n = 50 and p = 0.5
x
21
P( X ≤ x )
0.161118
the probability that less than 22 students in the sample graduate in 6 years is 0.161118
#5 Sketch of Solutions
#2
X = the number of alcohol related deaths in a random sample of 35 car crash deaths in Maine.
X is Binomial n = 35 and p = 0.29
Cumulative Distribution Function
Binomial with n = 35 and p = 0.29
x
14
P( X <= x )
0.943886
the probability that at most 14 were alcohol related is 0.943886
Cumulative Distribution Function
Binomial with n = 35 and p = 0.29
x
13
P( X <= x )
0.891948
the probability that fewer than 14 were alcohol related is 0.891948
Cumulative Distribution Function
Binomial with n = 35 and p = 0.29
x
22
P( X <= x )
0.99999
Cumulative Distribution Function
Binomial with n = 35 and p = 0.29
x
3
P( X <= x )
0.0034887
the probability that between 4 and 22, inclusive, were alcohol related is 0.99999 – 0.0034887 = 0.9965013
it would be surprising to find more than 65% (X > 22.75) of the sample were alcohol related –
since a very high likelihood of being between 4 and 22 inclusive …
#5 Sketch of Solutions
#3
Under no preference the distribution for the number of the 16 infants who choose the helper toy is
Binomial n = 16 p = 0.5
Cumulative Distribution Function
Binomial with n = 16 and p = 0.5
x
13
P( X <= x )
0.997910
P(X >= 14) = 1 – P(X < 14) = 1 – 0.997910 = 0.00209
This is a very low chance … if no preference - expect to see something much closer to 8 = 16(0.5) …
Therefore there is strong evidence to doubt the “no preference” assumption …
Indicating that infants really do show a preference for the helper toy …
#5 Sketch of Solutions
#4 TEXT page 264 #5.125
X = alkalinity level (mg/liter) of water from Han River = Seoul, Korea
mean = 50 and stdev = 3.2 – normally distributed
Cumulative Distribution Function
Normal with mean = 50 and standard deviation = 3.2
x
45
P( X ≤ x )
0.0590851
P(X > 45) = 1 – 0.0590851 = 0.9409149 = probability of exceeding 45 milligrams per liter.
Cumulative Distribution Function
Normal with mean = 50 and standard deviation = 3.2
x
55
P( X ≤ x )
0.940915
P(X < 55) = 0.940915 = probability of below 55 milligrams per liter.
Cumulative Distribution Function
Normal with mean = 50 and standard deviation = 3.2
x
52
P( X ≤ x )
0.734014
Cumulative Distribution Function
Normal with mean = 50 and standard deviation = 3.2
x
51
P( X ≤ x )
0.622670
P(51 <= X <= 52) = 0.734014-0.622670 = 0.111344 = probability b/w 51 and 52 milligrams per liter.
#5 Sketch of Solutions
#5 TEXT page 265 #5.133
X = hours of sleep daily – visually impaired students mean = 9.06 and stdev = 2.11 – normally distributed
Cumulative Distribution Function
Normal with mean = 9.06 and standard deviation = 2.11
x
6
P( X ≤ x )
0.0734962
P(X < 6) = 0.0734962 = probability of less than 6 hours of sleep.
Cumulative Distribution Function
Normal with mean = 9.06 and standard deviation = 2.11
x
10
P( X ≤ x )
0.672020
Cumulative Distribution Function
Normal with mean = 9.06 and standard deviation = 2.11
x
8
P( X ≤ x )
0.307704
P(8 <= X <= 10) = 0.672020-0.307704 = 0.364316 probability of b/w 8 and 10 hours of sleep.
Inverse Cumulative Distribution Function
Normal with mean = 9.06 and standard deviation = 2.11
P( X ≤ x )
0.2
x
7.28418
20% of visually impaired students obtain less than 7.28418 hours of sleep on a typical day.
#5 Sketch of Solutions
#6 TEXT page 244 #5.48
X = startle response(milliseconds) – mean = 37.9 and stdev = 12.4 – normally distributed …
Cumulative Distribution Function
Normal with mean = 37.9 and standard deviation = 12.4
x
50
P( X <= x )
0.835420
Cumulative Distribution Function
Normal with mean = 37.9 and standard deviation = 12.4
x
40
P( X <= x )
0.567241
probability b/w 40 and 50 milliseconds = 0.835420 – 0.567241 = 0.268179
Cumulative Distribution Function
Normal with mean = 37.9 and standard deviation = 12.4
x
30
P( X <= x )
0.262031
probability less than 30 milliseconds = 0.262031
Inverse Cumulative Distribution Function
Normal with mean = 37.9 and standard deviation = 12.4
P( X <= x )
0.975
x
62.2036
Inverse Cumulative Distribution Function
Normal with mean = 37.9 and standard deviation = 12.4
P( X <= x )
0.025
x
13.5964
middle 95% = 13.6 to 62.2 milliseconds
Inverse Cumulative Distribution Function
Normal with mean = 37.9 and standard deviation = 12.4
P( X <= x )
0.9
x
53.7912
10% above – 90th percentile = 53.8 milliseconds
#5 Sketch of Solutions
#7 TEXT page 243 #5.45
X = ambulance response time for EMS station Edmonton, Alberta == mean 7.5 min, stdev 2.5 min – normal
Inverse Cumulative Distribution Function
Normal with mean = 7.5 and standard deviation = 2.5
P( X ≤ x )
0.9
x
10.7039
Cumulative Distribution Function
Normal with mean = 7.5 and standard deviation = 2.5
x
9
P( X ≤ x )
0.725747
Regulations are NOT met – 90th percentile is 10.7 minutes and the % under 9 minutes is only 72.6%
Cumulative Distribution Function
Normal with mean = 7.5 and standard deviation = 2.5
x
2
P( X ≤ x )
0.0139034
2 minutes is 2.2 stdev below the mean and is in the extreme left tail (below the 2nd percentile).
Unlikely that the call was serviced by Station A.