Question

In: Statistics and Probability

2. Following is the weight data (in grams) of 25 tea bags samples produced by a...

2. Following is the weight data (in grams) of 25 tea bags samples produced by a machine in 1 hour: 5.65 5.44 5.42 5.40 5.53 | 5.34 5.54 5.45 5.52 5.41 | 5.57 5.40 5.53 5.54 5.55 | 5.62 5.56 5.46 5.44 5.51 | 5.47 5.40 5.47 5.61 5.53 |
a. Calculate the average, median, quartile 1 and quartile 3
b. Calculate range, interquartile range, variance, standard deviation, and coefficient of variation.
c.Draw the boxplot, give a conclusion

Solutions

Expert Solution

Solution:- given that 5.65,5.44,5.42,5.40,5.53,5.34,5.54,5.45,5.52,5.41,5.57,5.40,5.53,5.54,5.55,5.62,5.56,5.46,5.44,5.51,5.47,5.40,5.47,5.61,5.53

a. The aveage mean = sum of terms/number of terms
= 137.36/25
= 5.4944

b. The median of the data set is 5.51.

Explanation

The median is the middle number in a sorted list of numbers. So, to find the median, we need to place the numbers in value order and find the middle number.

Ordering the data from least to greatest, we get:

5.34 5.40 5.40 5.40 5.41 5.42 5.44 5.44 5.45 5.46 5.47 5.47 5.51 5.52 5.53 5.53 5.53 5.54 5.54 5.55 5.56 5.57 5.61 5.62 5.65   

So, the median is 5.51 .


The first quartile of the data set is 5.43.

Explanation

The first quartile (or lower quartile or 25th percentile) is the median of the bottom half of the numbers. So, to find the first quartile, we need to place the numbers in value order and find the bottom half.

5.34 5.40 5.40 5.40 5.41 5.42 5.44 5.44 5.45 5.46 5.47 5.47 5.51 5.52 5.53 5.53 5.53 5.54 5.54 5.55 5.56 5.57 5.61 5.62 5.65   

So, the bottom half is

5.34 5.40 5.40 5.40 5.41 5.42 5.44 5.44 5.45 5.46 5.47 5.47   

The median of these numbers is 5.43.

                                                                                                                             
The third quartile of the data set is 5.545.

Explanation

The third quartile (or upper quartile or 75th percentile) is the median of the upper half of the numbers. So, to find the third quartile, we need to place the numbers in value order and find the upper half.

5.34 5.40 5.40 5.40 5.41 5.42 5.44 5.44 5.45 5.46 5.47 5.47 5.51 5.52 5.53 5.53 5.53 5.54 5.54 5.55 5.56 5.57 5.61 5.62 5.65   

So, the upper half is

5.52 5.53 5.53 5.53 5.54 5.54 5.55 5.56 5.57 5.61 5.62 5.65   

The median of these numbers is 5.545.

b.

The range of data set is 0.31.

Explanation

The range is the difference between the highest and lowest values in the data set.

Ordering the data from least to greatest, we get:

5.34 5.40 5.40 5.40 5.41 5.42 5.44 5.44 5.45 5.46 5.47 5.47 5.51 5.52 5.53 5.53 5.53 5.54 5.54 5.55 5.56 5.57 5.61 5.62 5.65   

The lowest value is 5.34.

The highest value is 5.65.

The range = 5.65 - 5.34 = 0.31.

The interquartile range of the data set is 0.115.

Explanation

The interquartile range is the difference between the third and first quartiles.

The third quartile is 5.545.

The first quartile is 5.43.

The interquartile range = 5.545 - 5.43 = 0.115.

=> variance = 0.0062

=> standard deviation = sqrt(variance) = sqrt(0.0062) = 0.0787

=> Coefficient of variance = sd/mean = 0.0143

c. box plot :-



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