Question

Question 7

1. The mean of a distribution is 50 and the standard deviation is 5. Between what to values will at least 75% of the data values fall? Blank 1 and Blank 2

Question 8

1. Twelve secretaries were given a typing test, and the times (in minutes) to completed it were as follows: 8, 12, 15, 9, 6, 8, 10, 9, 8, 6, 7, 8 Find the MEDIAN.

Question 9

1. Twelve secretaries were given a typing test, and the times (in minutes) to completed it were as follows: 8, 12, 15, 9, 6, 8, 10, 9, 8, 6, 7, 8 Find the VARIANCE. Round to one decimal place (example: 3.2)

  

Question 10

1. The mean of a distribution is 50 and the standard deviation is 5. Between what to values will at least 88.89% of the data values fall? Blank 1 and Blank 2

Answer 1

Question 7:

Given mean 50

standard deviation 5

using Chebyshev's theorem

1-(1 / k^2) = 75%

1-(1 / k^2) = 0.75

1 = 0.75 + 1/K^2

1-0.75 = 1/k^2

0.25 = 1/k^2

0.25*k^2 = 1

k^2 = 1/ 0.25

k^2 = 4

k = sqrt(4)

k = 2

at least 75% of the data values fall between

mean-k(standard deviation ) and mean+k(standard deviation)

50-2(5) and 50+2(5)

50-10 and 50+10

40 and 60.

QUESTION 8

GIVEN

8, 12, 15, 9, 6, 8, 10, 9, 8, 6, 7, 8

MEDIAN = middle number

listed in numerical order =6,6,7,8,8,8,8,9,9,10,12,15

MEDIAN =( 8+8 )/2 = 16/2 = 8

QUESTION 9

Given 8, 12, 15, 9, 6, 8, 10, 9, 8, 6, 7, 8

therefore variance = 6.5

Question 10

Given mean = 50

standard deviation = 5

At least 88.89% of the data values fall between 1.6 sigma according to the 3-sigma rule.

Hence C.I = (50 +- 1.6 x 5) = (50 +- 8) =( 50-8 , 50=8) = (42 , 58 )