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Frequency Distributions - Assignment Example

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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?
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Frequency Distributions
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Case Assignment Answer the following problems showing your work and explaining (or analyzing) your results 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   89Ans: The mean is 77.93 and the Standard deviation is 18.38.

There were 12 observations within one Standard Dev of mean limitsThere were 14 observations within two Standard Dev of mean limits 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%?Ans: Mean=74.8 and S.D= 7.57, then z= (85-74.8) / 7.57 i.e 1.3474Cumulative area for z score 1.3474 is 0.911 i.e. cumulative area for z score>1.3474 is (1-0.911) = 0.089.Therefore student scoring above 85% is 15*0.

089 which is 1.335 or 1 (approx).3. If you know the standard deviation, how do you find the variance? Ans: If we square the Standard Deviation we get Variance.Variance= (Standard Deviation)^24. 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   $235Find the standard deviation.Ans: Standard Deviation is 48.885. A company has 70 employees whose salaries are summarized in the frequency distribution below.

SalaryNumber of Employees5,001–10,000810,001–15,0001215,001–20,0002020,001–25,0001725,001–30,00013a. Find the standard deviation.b. Find the variance.Ans: standard deviation is 1.252Variance is 1.5686.  Calculate the mean and variance of the data. Show and explain your steps. Round to the nearest tenth.       14,   16,   7,   9,   11,   13,   8,   10Ans: Mean is 11 and Variance is 8.5.The steps are shown below:a) calculate the mean i.e sum(all obsv)/no.of obsv.(14+16+7+9+11+13+8+10) / 8b) calculate ( obsv-mean )^2 as shown belowc) Calculate sum of (obsv-mean) ^2 and divide it by no of obsv to get the variance. i.e. 68/87. Create a frequency distribution table for the number of times a number was rolled on a die.

(It may be helpful to print or write out all of the numbers so none are excluded.)    3,   5,   1,   6,   1,   2,   2,   6,   3,   4,   5,   1,   1,   3,   4,   2,   1,   6,   5,   3,   4,   2,   1,   3,   2,   4,   6,   5,   3,   1Ans: Frequency distribution is shown below:8. Answer the following questions using the frequency distribution table you created in No. 7.a. Which number(s) had the highest frequency?b. How many times did a number of 4 or greater get thrown?c. How many times was an odd number thrown?d. How many times did a number greater than or equal to 2 and less than or equal to 5 get thrown?

Ans: The table below gives the answer: 9. The wait times (in seconds) for fast food service at two burger companies were recorded for quality assurance. Using the data below, find the following for each sample.a. Rangeb. Standard deviationc. VarianceLastly, compare the two sets of results.Ans: The answer is given below in table: CompanyWait times in secondsBig Burger Company105677812017511512059The Cheesy Burger1331242007910114711812510. What does it mean if a graph is normally distributed? What percent of values fall within 1, 2, and 3, standard deviations from the mean?

Ans: A graph is said to be normally distributed if it is symmetric and bell shaped with a single peak. For any normal distribution two quantities must be defined, the mean “mu” where the peak of the curve occurs and standard deviation “sigma” which signifies the spread of the curve from its mean value.68% of the observations fall within 1 standard deviation of the mean.95% of the observations fall within 2 standard deviation of the mean.99.7% of the observations fall within 3 standard deviation of the mean.

ReferenceDas, N.G. (2006). Statistical Methods. Kolkata: Word Press

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