Question

Hi!

Can someone please help me with the following question? If you can show all your steps so I can understand the solution I would really appreciate it!
Thanks :)

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2- Below you can find scores of a test: {56,67,67,70,73,73,75,76,77,79,80,81,83,83,85,87,88,89,92,93 } (a) Obtain a frequency distribution for the test. (The number of bins and intervals should be determined according to the rules in the class slides). (b) Draw a frequency histogram. (c) Use the frequency distribution to compute the mean and the median and variance. Are there any outliers in this data set? (d) Plot the box plot of data based on the statistics you can obtain from raw data. (e) Calculate the standard score for the test scores of 56 and 92 based on the statistics you obtained from (c). (1) With assumption that professor's ideal distribution has a mean of 70 and standard deviation of 10, calculate new score of students (curving the grades) with the following original scores {56,70, 92}.

Answer 1

2.

(a)

We take classes as 50-59, 60-69, 70-79, 80-89, 90-99 as so frequency distribution is as follows.

Class limits	Class boundaries	Class mark (xi)	Frequency (fi)
56-60	55.5-60.5	58	1
61-65	60.5-65.5	63	0
66-70	65.5-70.5	68	3
71-75	70.5-75.5	73	3
76-80	75.5-80.5	78	4
81-85	80.5-85.5	83	4
86-90	85.5-90.5	88	3
91-95	90.5-95.5	93	2
Total	-		20

(b)

The frequency histogram is as follows.

Histogram of x 4 3 Frequency 2 60 70 80 90

(c)

Mean is given by

- Σ= fiti - 1575 = 78,75

Median is the 20/2 = 10th element in increasing order. So, it is the 3rd element of the class 76-80 (median class) and so median can be obtain as follows.

T =1+ 1/2-C *h

where

l = lower limit of median class interval = 76

C= cumulative frequency of preceding to the median class frequency = 1+0+3+3 = 7

f = frequency of the class interval to which median belongs = 4

h = width of the class interval = 80.5-75.5 = 5

N= total frequency = 20

...0 = 76 + 10 – 7 -*5= 79.75

Variance is given by

2 Σ= f(x; – 1)? _ 1825 - 91.25 20

(d)

Box plot is as follows.

160 7 0 80 90

(e)

For 56, standard score is given by

56 – 78.75 -= -2.381579 91.25

For 92, standard score is given by

92 – 78.75 -= 1.387073 91.25

(f)

Mean = 70 and standard deviation = 10

For 56, standard score is given by

56 – 70 -1.4 10

For 70, standard score is given by

70-70 - = 0 10

For 92, standard score is given by

92 – 70 102.2