1. A fertilizer mixing machine is set to give 12kg of nitrate for every 100Kg bag of fertilizer. A sample of ten such bags is examined. The percentages of nitrates are as follows:
11, 14, 13, 12, 13, 12, 13, 14, 11, 12.
a. Calculate
i. The sample mean
ii. Sample standard deviation s
Answers
Answer:
Step-by-step explanation:
Missing point in question :- Is there any reason to believe that the machine is
defective?
Answer:-
If the probability of bags ≠ 12 kg is > 5% , then there is every reason to believe that the machine is defective.
Out of 10 samples:- 11, 14, 13, 12, 13, 12, 13, 14, 11, 12 only 3 , 12 kg bags are seen in sample. So probability of bags ≠ 12 kg = 70%
Now, the sample size was only 10 bags out of 100. So the probability when extended to 100 bags become = 1/10 x 70% = 7%
Since 7% > 5% , then there is every reason to believe that the machine is defective
Given:
The percentages of nitrates in 10 bags of fertilizers- 11, 14, 13, 12, 13, 12, 13, 14, 11, 12
To find:
The sample mean and sample standard deviation
Solution:
We can find the values by following the given steps-
We know that the sample mean is calculated by dividing the sum of terms by the total number of terms.
Total number of terms=10
The sum of percentages of all the fertilizers= 11+14+13+12+13+12+13+14+11+12
=25+25+25+27+23
=75+50
=125
The sample mean= Sum of percentages/ Number of percentages
=125/10
Mean=12.5
The value of standard deviation is obtained by taking the root of the sum of the difference of all terms and the sample mean divided by the number of terms.
The difference of terms and mean is as follows-
11-12.5= -1.5
14-12.5=1.5
13-12.5=0.5
12+12.5= -0.5
13-12.5=0.5
12+12.5= -0.5
13-12.5=0.5
14-12.5=1.5
11-12.5= -1.5
12+12.5= -0.5
On squaring and adding these, we get
=(-1.5)²+1.5²+0.5²+(-0.5)²+0.5²+(-0.5)²+0.5²+1.5²+(-1.5)²+(-0.5)²
= 2.25+2.25+0.25×5+2.25+2.25+0.25
=9+0.25×6
=9+1.5
=10.5
The standard deviation=√10.5/√10
s=1.02
Therefore, the sample mean is 12.5, and the sample standard deviation is 1.02.