Using the frequency table. what fraction of students passed the test
1 answer:
You might be interested in
Answer is explained below
The average annual temperatures (°F) from 1991 to 2000 for Kentucky were 57.5, 55.1, 52.2, 55.6, 55.6, 54.4, 54.5, 58.3, 57.3, and 55.7.
We have to evaluate the measures of center.
Firstly, we will calculate the mean.
Mean = (Sum of all the observations)/ (Total number of observations)
Mean =
=
= 55.62
Now, we will calculate mode.
Mode is a statistical term that refers to the most frequently occurring number found in a set of numbers.
Consider the data 57.5, 55.1, 52.2, 55.6, 55.6, 54.4, 54.5, 58.3, 57.3, and 55.7.
Mode = 55.6
Now, we will calculate the median.
Firstly, we will arrange in the ascending order.
52.2, 54.4, 54.4,55.1 ,55.6, 55.6, 55.7, 57.3, 57.5, 58.3
Since there are 10 terms.
Median = th observation
Median = 5th observation.
Since, the 5th observation is 55.6.
Therefore, the median is 55.6