BUS 219 / MAT 167 Summer, 2012 Final Exam 300 points # Please show all pertinent steps. A top-notch solution is complete and is easy to # follow from beginning to end. The weights of first graders at...

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Answer To: BUS 219 / MAT 167 Summer, 2012 Final Exam 300 points # Please show all pertinent steps. A top-notch...

David answered on Dec 29 2021
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BUS 219
BUS 219 / MAT 167
Summer, 2012
Final Exam
300 points
# Please show all pertinent steps. A top-notch solution is complete and is easy to # follow from beginning to end.
1. The weights of first graders at Freemont Elementary were recorded as follows. Construct
a frequency distribution (including cumulative frequency) using 6 classes and starting with 45.
45
50
60
65
62
57
62
65
49
46
50
51
58
62
67
71
    classes
    Frequency
    45-50
    5
    50-55
    2
    55-60
    3
    60-65
    4
    65-70
    1
    >70
    1
2. Find the range and variance of the following data set:
45, 51, 53, 76
Range= 76-45= 31
Mean =(45+51+53+76)/4 = 56.25
Var= ( 45-56.25) ^2 + ( 51-56.25) ^2 +( 53-56.25) ^2 +( 76-56.25) ^2 =554.75
3. Find the mean and variance of the probability distribution:
,012
131
Pr,()
882
NumberofGirlsX
obabilityPX
    X
    P(X)
    X*P(X)
    X*X*P(X)
    0
    0.125
    0
    0
    1
    0.375
    0.375
    0.375
    2
    0.5
    1
    2
     
     
     
     
     
     
    1.375
    2.375
Mean= 1.375 and var= 2.375 –(1.375)^2 = 0.484375
4. Compute the binomial probability if n = 21, p = 0.42, and x = 10
Prob = 21C10 .4210 .5811 = 0.15053101634159 = .1505
5. A.C. Neilson reported that children between the ages of 4 and 9 watch an average of 22 hours of TV per week. Assume the variable is normally distributed and the standard deviation is 4 hours. find the probability that a randomly selected child will watch fewer than 18 hours of TV.
Let X be the no of hours that children between the ages of 4 and 9 watch TV
P( X< 18) = P( z<( 18-22)/4) = P( z <-1) = .15866
The prob can be computed from http://stattrek.com/online-calculator/normal.aspx
6. The average teacher’s salary in Arizona is $42,300 with standard deviation $5000. Assuming a normal distribution, what is the probability that the mean of a sample of 50 teachers’ salaries will be at least $48,000 a year?
Let X be the salary of teachers
We use the Central Limit theorem that says that the sample mean is normally distributed with mean equal to population mean and standard deviation equal to std dev of population/ n.5
Mean of sample mean= Population mean = 42300
Std dev of sample mean = 5000/50^.5
P ( sample X > 48000) = P( z > (48000-42300)*50.5/5000)= P(z <>8.06) =0
7. If a baseball player's batting average is 0.275, find the probability that the player will get at least 40 hits in 100 times at bat.
This is a binomial distribution with n=100 p= .275
Np=27.5 > 5 so it can be approximated by normal distribution.
Mean = 27.5
Variance = 27.5*.725= 19.9375
P( X > 40) = P( z > (40-27.5)/19.9375.5) = P(z >2.799) = .00256
8. Thirty randomly selected vehicles were stopped, and the tread depth of the right front tire was measured. The mean was 0.20 inch, and the standard...
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