Question

1) - On a final examination in mathematics, the mean was 72 and the variance was 225. Determine the standard scores (u.e., grades in standard-deviation units) of students receiving the grades (a) 60, (b) 93, and (c) 72.

2) - Find the Mean, Median, and Mode for the sets (a) 3, 5, 2, 6, 5, 9, 5, 2, 8, 6 and (b) 51.6, 48.7, 50.3, 49.5, 48.9

Answer 1

Solution :

Given that ,

1) mean = = 72

variance = ² = 225

standard deviation = = ² = 225 = 15

a) x = 60

Using z-score formula,

z = x - /   

z = 60 - 72 / 15

z = -0.80

standard score = z = -0.80

b) x = 93

Using z-score formula,

z = x - /   

z = 93 - 72 / 15

z = 1.40

standard score = z = 1.40

c) x = 72

Using z-score formula,

z = x - /   

z = 72 - 72 / 15

z = 0.00

standard score = z = 0.00

2) a) 3, 5, 2, 6, 5, 9, 5, 2, 8, 6

Arrange the data set lowest to highest

2,2,3,5,5,5,6,6,8,9

n = 10

The mean of sample is   

x/n =  (2 + 2 + 3 + 5 + 5 + 5 + 6 + 6 + 8 + 9 / 10)

   = 51/ 10

mean = 5.1

Median = Median is middle most observation of data set,

2,2,3,5,5,5,6,6,8,9

Median = 5 + 5 / 2

  Median = 5

Mode = Represent the repeated value in data set,

   2,2,3,5,5,5,6,6,8,9

Mode = 5

(b) 51.6, 48.7, 50.3, 49.5, 48.9

Arrange the data set lowest to highest

48.7, 48.9, 49.5, 50.3, 51.6

n = 5

The mean of sample is   

x/n =  (48.7 + 48.9 + 49.5 + 50.3 + 51.6 / 5)

   = 249 / 5

mean = 49.8

Median = Median is middle most observation of data set,

48.7, 48.9, 49.5, 50.3, 51.6

Median = 49.5

Mode = not existent