The twenty units of potatoes illustrates the measure of central tendency and probability
The modal weight, mean, and median
Start by sorting the dataset:
0.19; 0.28; 0.55;
0.62; 0.68; 0.77;
0.88; 0.89; 0.98;
0.99; 1.11; 1.22;
1.22; 1.22; 1.33;
1.44; 1.48; 1.68;
1.77; 1.99
The highest occuring element is 1.22 with a frequency of 3
The two middle elements are 0.99 and 1.11.
The mean of these two middle elements is: 0.5 * (0.99 + 1.11) = 1.05
The mean of the whole dataset is:
Mean = (0.19+ 0.28+ 0.55+ 0.62+ 0.68+ 0.77+ 0.88+ 0.89+ 0.98+ 0.99+ 1.11+ 1.22+1.22+ 1.22+ 1.33+ 1.44+ 1.48+ 1.68+1.77+1.99)/20
Mean = 1.06
Hence, the modal weight is 1.22, the mean is 1.06, and the median is 1.05
The probabilities
(a) at least 1.1 kg
There are 10 potatoes that weigh at least 1.1 kg.
So, the probability is:
P = 10/20
P = 0.5
(b) more than 3.5kg
There are 0 potatoes that weigh more than 3.5 kg.
So, the probability is:
P = 0/20
P = 0
(c) below 0.50kg
There are 2 potatoes that weigh below 0.50 kg.
So, the probability is:
P = 2/20
P = 0.1
The Variance of the weight.
This is calculated as:
Var = [tex]\sum[/tex](x - mean)^2/n
This gives
Var = [(0.19 - 1.06)^2 + (0.28 - 1.06)^2 + (0.55 - 1.06)^2 + (0.62 - 1.06)^2 + (0.68 - 1.06)^2 + (0.77 - 1.06)^2 + (0.88 - 1.06)^2 + (0.89 - 1.06)^2 + (0.98 - 1.06)^2 + (0.99 - 1.06)^2 + (1.11 - 1.06)^2 + (1.22 - 1.06)^2 + (1.22 - 1.06)^2 + (1.22 - 1.06)^2 + (1.33 - 1.06)^2 + (1.44 - 1.06)^2 + (1.48 - 1.06)^2 + (1.68 - 1.06)^2 + (1.77 - 1.06)^2 + (1.99 - 1.06)^2]/20
Evaluate
Var = 0.22
This means that:
The expectation of the squared deviation of the random variable from the mean is 0.22 kg
Read more about measure of central tendency at:
https://brainly.com/question/17631693
