Frequency Distributions

Frequency distributions
Name
Course
Instructor
Date
[1] The math grades on the final exam varied greatly. Using the scores below, how many scores were within one standard deviation of the mean? How many scores were within two standard deviations of the mean?
99 34 86 57 73 85 91 93 46 96 88 79 68 85 89
xx-u(x-u)^29921.0667443.80534-43.9331930.14868.0666765.071257-20.933438.20473-4.933324.3377857.0666749.93789113.0667170.7389315.0667227.00546-31.9331019.749618.0667326.4058810.0667101.338791.066671.1377868-9.933398.671857.0666749.93788911.0667122.4715068.93Population variance(5068.933/15)337.929Sample variance(5068.993/14)362.067Population standard deviation18.3828Sample standard deviation19.028
Mean ±1 σ= 77.93333 ±1 (18.3828) = 59.5505- 96.3161
Rearranged data set 34,46,57,68,73,79,85,85,86,88,89,91,93,96,99
Mean ±1 σ= 77.93333 ±2 (18.3828) = 41.1677- 114.699
Answer
Hence, there are 11 numbers within 1 standard deviation
Hence, there are 14 numbers within 2 standard deviations
[2] The scores for math test #3 were normally distributed. If 15 students had a mean score of 74.8% and a standard deviation of 7.57, how many students scored above an 85%?
From the Z table at 85% confidence table the score is (1-0.85)/2=0.075
The Z score is (85-74.8)/7.57= 1.34742
Score from the z table 0.088922
15*= 0.088922=1.3, only 1 student scored above 85%
[3] If you know the standard deviation, how do you find the variance?
Since the standard deviation is the square root of the variance, to get the variance one needs to square the standard deviation. The standard deviation represents how the data values are spread since it is a measure of variability (Lane, n.d.).
[4] To get the best deal on a stereo system, Louis called eight appliance stores and asked for the cost of a specific model. The prices he was quoted are listed below:
$216   $135   $281   $189   $218   $193   $299   $235
Sum= 1766
Mean=1766/8=220.75
xx-u(x-u)^22164.7522.562513585.757353.06281-60.253630.0618931.751008.062182.757.562519327.75770.063299-78.256123.06235-14.25203.063176619117.5Population variance2389.69Sample variance2731.07Population standard deviation48.8844Sample standard deviation52.2596[5]A company has 70 employees whose salaries are summarized in the frequency distribution below.
SalaryNumber of Employees5,001–10,000810,001–15,0001215,001–20,0002020,001–25,0001725,001–30,00013
SalaryNumber of EmployeesMidpoint Fxx^2Fx^25,001–10,00087500.56000456257500.2545006000210,001–15,0001212500.5150006156262500.3187515000315,001–20,0002017500.5350010306267500.3612535000520,001–25,0001722500.5382508.5506272500.3860663250425,001–30,0001327500.5357506.5756277500.39831607503130003526888800018
S.D=‚ [26888800018/70-(13...