88. Refer to the Baseball 2005 data, which reports information on the 30 major league teams for the 2005 baseball season.
a. Select the variable team salary and find the mean, median, and the standard deviation.
b. Select the variable that refers to the age the stadium was built. (Hint: Subtract the
year in which the stadium was built from the current year to find the stadium age and
work with that variable.) Find the mean, median, and the standard deviation.
c. Select the variable that refers to the seating capacity of the stadium. Find the mean,
median, and the standard deviation.
Solution:
a. Select the variable team salary and find the mean, median, and the standard deviation.
Team / Salary( in Million Dollars)Boston / 123.5
New York Yankees / 208.3
Oakland / 55.4
Baltimore / 73.9
Los Angeles Angels / 97.7
Cleveland / 41.5
Chicago White Sox / 75.2
Toronto / 45.7
Minnesota / 56.2
Tampa Bay / 29.7
Texas / 55.8
Detroit / 69.1
Seattle / 87.8
Kansas City / 36.9
Atlanta / 86.5
Arizona / 62.3
Houston / 76.8
Cincinnati / 61.9
New York Mets / 101.3
Pittsburgh / 38.1
Los Angeles Dodgers / 83
San Diego / 63.3
Washington / 48.6
San Francisco / 90.2
St. Louis / 92.1
Florida / 60.4
Philadelphia / 95.5
Milwaukee / 39.9
Chicago Cubs / 87
Colorado / 48.2
Mean / $73.0600
Median / 66.2
Standard Deviation / 34.23285
b. Select the variable that refers to the age the stadium was built. (Hint: Subtract the
year in which the stadium was built from the current year to find the stadium age and
work with that variable.) Find the mean, median, and the standard deviation.
Team / Built Age / Age (2009-Built Age)Boston / 1912 / 97
New York Yankees / 1923 / 86
Oakland / 1966 / 43
Baltimore / 1992 / 17
Los Angeles Angels / 1966 / 43
Cleveland / 1994 / 15
Chicago White Sox / 1991 / 18
Toronto / 1989 / 20
Minnesota / 1982 / 27
Tampa Bay / 1990 / 19
Texas / 1994 / 15
Detroit / 2000 / 9
Seattle / 1999 / 10
Kansas City / 1973 / 36
Atlanta / 1993 / 16
Arizona / 1998 / 11
Houston / 2000 / 9
Cincinnati / 2003 / 6
New York Mets / 1964 / 45
Pittsburgh / 2001 / 8
Los Angeles Dodgers / 1962 / 47
San Diego / 2004 / 5
Washington / 1961 / 48
San Francisco / 2000 / 9
St. Louis / 1966 / 43
Florida / 1987 / 22
Philadelphia / 2004 / 5
Milwaukee / 2001 / 8
Chicago Cubs / 1914 / 95
Colorado / 1995 / 14
Mean / 28.2
Median / 17.5
Standard Deviation / 25.94477
c. Select the variable that refers to the seating capacity of the stadium. Find the mean,
median, and the standard deviation.
Team / Seating Capacity of the StadiumBoston / 33871
New York Yankees / 57746
Oakland / 43662
Baltimore / 48262
Los Angeles Angels / 45050
Cleveland / 43368
Chicago White Sox / 44321
Toronto / 50516
Minnesota / 48678
Tampa Bay / 44027
Texas / 52000
Detroit / 40000
Seattle / 45611
Kansas City / 40529
Atlanta / 50062
Arizona / 49075
Houston / 42000
Cincinnati / 42059
New York Mets / 55775
Pittsburgh / 38127
Los Angeles Dodgers / 56000
San Diego / 42445
Washington / 56000
San Francisco / 40800
St. Louis / 49625
Florida / 42531
Philadelphia / 43500
Milwaukee / 42400
Chicago Cubs / 38957
Colorado / 50381
Mean / 45912.60
Median / 44174.00
Standard Deviation / 5894.20
56. Assume the likelihood that any flight on Northwest Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study.
a. What is the likelihood all four of the selected flights arrived within 15 minutes of the
scheduled time?
b. What is the likelihood that none of the selected flights arrived within 15 minutes of
the scheduled time?
c. What is the likelihood at least one of the selected flights did not arrive within 15 minutes
of the scheduled time?
Solution:
Let X be the sample parameter for any flight on Northwest Airlines arrives within 15 minutes of
the scheduled time.
Sample of size, n = 4
p = 0.9.
q= 0.1
a. P(X=4) = 0.94 = 0.6561
b. P(X=0) = (1-0.9)4 = 0.0001
c. P (X≤3) = 1 - 0.6561 = 0.3439 since P (X≤3) = 1- P(X=4)
64. An internal study by the Technology Services department at Lahey Electronics revealed
company employees receive an average of two emails per hour. Assume the arrival of
these emails is approximated by the Poisson distribution.
a. What is the probability Linda Lahey, company president, received exactly 1 email
between 4 P.M. and 5 P.M. yesterday?
b. What is the probability she received 5 or more email during the same period?
c. What is the probability she did not receive any email during the period?
Solution:
where
· e is the base of the natural logarithm (e = 2.71828...)
· k is the number of occurrences of an event - the probability of which is given by the function
· k! is the factorial of k
· λ is a positive real number, equal to the expected number of occurrences that occur during the given interval. For instance, if the events occur on average 4 times per minute, and you are interested in the number of events occurring in a 10 minute interval, you would use as your model a Poisson distribution with λ=10×4=40.
(Source: http://en.wikipedia.org/wiki/Poisson_distribution)
(a) P(received exactly 1 email between 4 P.M. and 5 P.M. yesterday)=f(1,2) = 2e-2 = 0.2707
(b)P(received 5 or more email during the same period) =1-k=0∑k=4f(k;λ)
= 1-[(e-2+2e-2+2e-2+(4/3)e-2+(2/3)e-2] = 0.05265
(c) P(she did not receive anyemail during period) = f(0,2)= e-2 = 0.13534
50. Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made
a study of the maintenance costs and determined the number of miles traveled during
the year followed the normal distribution. The mean of the distribution was 60,000 miles
and the standard deviation 2,000 miles.
a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more?
b. What percent of the trucks logged more than 57,060 but less than 58,280 miles?
c. What percent of the Fords traveled 62,000 miles or less during the year?
d. Is it reasonable to conclude that any of the trucks were driven more than 70,000 miles?
Explain.
Solution:
a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more?
Here, sample mean is 65,200 miles and population mean is 60,000 miles.
the standard deviation is 2,000 miles.
(a) Calculation of test-statistics,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σx is the standard deviation.
Hence, z= 65,200-60,000/2,000 = 5,200/2000 = 2.6
P (the Ford Super Duty F-750s logged 65,200 miles or more) = P(Z>2.6)=1-P(Z<2.6)=1-0.9953=0.0047
Hence answer is 0.47%
b. What percent of the trucks logged more than 57,060 but less than 58,280 miles?
For X = 57,060,
Z = 57,060-60,000/2,000= -1.47
For X=58,280
Z= 58,250-60,000/2,000 = -0.875
P(the trucks logged more than 57,060 but less than 58,280 miles) = P(-1.47<Z<-0.875)
=P(Z<-0.875)-P(Z<-1.47) =0.1908-0.0708=0.12
c. What percent of the Fords traveled 62,000 miles or less during the year?
For X = 62000,
Z = 62,000-60,000/2,000 = 1
P(Fords traveled 62,000 miles or less during the year) = =P(Z<1)=0.8413
Hence, percentage is 84.13%
d. Is it reasonable to conclude that any of the trucks were driven more than 70,000 miles? Explain.
For X = 70000,
Z = 70,000-60,000/2,000= 5
P(any of the trucks were driven more than 70,000 miles
= P(Z>5) -> 0
Hence, it is not reasonable to assume that any of the trucks were driven more than 70,000 miles.
38. The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50 with a standard deviation of $5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions.
a. What is the likelihood the sample mean is at least $25.00?
b. What is the likelihood the sample mean is greater than $22.50 but less than $25.00?
c. Within what limits will 90 percent of the sample means occur?
Solution:
Here, population mean is $23.50
Standard Deviation σ = 5
n= size of sample = 50
Calculation of test-statistics,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
σx = σ/√n
σx = σ/sqrt(n) = 5/√50 = 0.7071
(a) What is the likelihood the sample mean is at least $25.00?
Calculation of test-statistics,
z= (25-23.5)/(5/√50) = 2.12
P(the sample mean is at least $25.00) = P(Z ≥ 2.12) = 0.017
(b) What is the likelihood the sample mean is greater than $22.50 but less than $25.00?
for sample mean = $22.50,
z = 22.5-23.5/(5/√50 = -1.41
P(the sample mean is greater than $22.50 but less than $25.00)
= P(-1.41 < Z < 2.12) = 0.9037
(c) Within what limits will 90 percent of the sample means occur?
90% Confidence Interval can be determined by μ - z*σ/√(n) to μ + z*σ/√(n).
Since it is 90% Confidence Interval, significant level will be 0.10.
Critical vaue of z for significant level 0.10 = 1.645
So, (23.5 - 1.645*5/√(50), 23.5 + 1.645*5/√(50))
= (22.33,24.67)
54. Families USA, a monthly magazine that discusses issues related to health and health
costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums
for a family with coverage through an employer averaged $10,979. The standard deviation
of the sample was $1,000.
a. Based on this sample information, develop a 90 percent confidence interval for the
population mean yearly premium.
b. How large a sample is needed to find the population mean within $250 at 99 percent
confidence?
Solution:
Here, population mean is $10,979
n= size of sample = 20
Calculation of test-statistics,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
σx = σ/sqrt(n)
a. Based on this sample information, develop a 90 percent confidence interval for the
population mean yearly premium.
90% Confidence Interval can be determined by μ - z*σ/sqrt(n) to μ + z*σ/sqrt(n).
Since it is 90% Confidence Interval, significant level will be 0.10.
Critical value of z for significant level 0.10 = 1.645
So, (10979-1.645*1000/√20, 10979+1.645*1000/√20)
= (10611.12,11346.81)
b. How large a sample is needed to find the population mean within $250 at 99 percent
confidence?
Since it is 99% Confidence Interval, significant level will be 0.01.
Critical value of z for significant level 0.01 = 2.5758
z = (X - μ) / σ/sqrt(n)
Now, z < (X - μ) / σ/sqrt(n)
Then z*σ/sqrt(n) < 250
=> 2.5758 * 1000/sqrt(n) < 250
=> 2.5758 * 1000/250 < sqrt(n)
=> 10.312 < sqrt(n)
=> 106.33 < n
This gives the sample size of 107.
42. During recent seasons, Major League Baseball has been criticized for the length of the
games. A report indicated that the average game lasts 3 hours and 30 minutes. A sample
of 17 games revealed the following times to completion. (Note that the minutes have
been changed to fractions of hours, so that a game that lasted 2 hours and 24 minutes
is reported at 2.40 hours.)
Solution:
Size of sample, n = 17
Degree of freedom = n-1 = 17-1 = 16
Sum of sample = i=1∑n=17xi = (2.98 + 2.40 + 2.70 + 2.25 + 3.23 + 3.17 + 2.93 + 3.18 +2.80
+2.38+ 3.75+ 3.20+ 3.27+ 2.52+ 2.58+ 4.45+ 2.45) = 50.24
1 / 2.98 / 50.24 / Mean = Sum/n
= 50.24/17
= 2.9552 / 0.000605 / σ = √( i=1∑n=25(xi-xm)2/(n-1))
= √(5.009623/16)
= √0.3131
=0.5595
2 / 2.4 / 0.308462
3 / 2.7 / 0.065226
4 / 2.25 / 0.497581
5 / 3.23 / 0.075408
6 / 3.17 / 0.046056
7 / 2.93 / 0.000645
8 / 3.18 / 0.050448
9 / 2.8 / 0.024147
10 / 2.38 / 0.331078
11 / 3.75 / 0.631399
12 / 3.2 / 0.059832
13 / 3.27 / 0.098977
14 / 2.52 / 0.189568
15 / 2.58 / 0.140921
16 / 4.45 / 2.233847
17 / 2.45 / 0.255423
Here X = 2.9552
σ = 0.5595
We can define one-tailed statistics for above observations as follows:
Null Hypothesis: H0: m =3.5 vs. Ha: m < 3.5
Rejection region:
t < -ta
Here significance level is 0.05,
So, t0.05 = - 2.120 (Using Statistical Ratio Calculator
From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating t with 0.05 significance and DF =16)
I am assuming the sample is selected independently and randomly from population. Population size is sufficiently large in sample.
Now,
t = (X - μ) / (s/√n)
Where X is a normal random variable, μ is the mean, and s is the standard deviation.