
How do you express the confidence interval $0.111 < p < 0.333$ in the form of $p\pm E$?
Answer
525.3k+ views
Hint: For this problem we need to write the given confidence interval in form of $p\pm E$. We know that the $p$ denotes the sample proportion and $E$ denotes the margin of error. First, we will calculate the margin of error and after that we will use this value to calculate the sample proportion. We know that the margin error is the half of the difference between the given limits. So, we will calculate the margin error for the given limits. Now we will use this value and consider either lower limit or upper limit to calculate the sample proportion. If you consider the upper limit, we will subtract the margin of error from the upper limit. If you consider the lower limit, we will add the margin of error to the lower limit to calculate the sample proportion. After getting both the values we will write it in required form.
Complete step by step solution:
Given confidence interval $0.111 < p < 0.333$.
The upper limit in the above interval is $0.333$.
The lower limit in the above interval is $0.111$.
Difference between the both the limits is $0.333-0.111=0.222$.
Half of the difference between the both the limits is $\dfrac{0.222}{2}=0.111$.
Hence the Margin of error is $E=0.111$.
Considering the lower limit of the given interval which is $0.111$.
To calculate the sample proportion, we are going to add the margin of error to the lower limit then we will get
$p=0.111+0.111=0.222$
Now the confidence interval in $p\pm E$ from is $0.222\pm 0.111$.
Note:
We can also calculate the sample proportion without calculating the margin of error. We have the formula for the sample proportion which is
$p=\dfrac{U.L+L.L}{2}$
So, we will calculate the average of the given limits and that will be our sample proportion.
Complete step by step solution:
Given confidence interval $0.111 < p < 0.333$.
The upper limit in the above interval is $0.333$.
The lower limit in the above interval is $0.111$.
Difference between the both the limits is $0.333-0.111=0.222$.
Half of the difference between the both the limits is $\dfrac{0.222}{2}=0.111$.
Hence the Margin of error is $E=0.111$.
Considering the lower limit of the given interval which is $0.111$.
To calculate the sample proportion, we are going to add the margin of error to the lower limit then we will get
$p=0.111+0.111=0.222$
Now the confidence interval in $p\pm E$ from is $0.222\pm 0.111$.
Note:
We can also calculate the sample proportion without calculating the margin of error. We have the formula for the sample proportion which is
$p=\dfrac{U.L+L.L}{2}$
So, we will calculate the average of the given limits and that will be our sample proportion.
Recently Updated Pages
The height of a solid metal cylinder is 20cm Its r-class-10-maths-ICSE

If a train crossed a pole at a speed of 60kmhr in 30 class 10 physics CBSE

Name the Writs that the High Courts are empowered to class 10 social science CBSE

A tower is 5sqrt 3 meter high Find the angle of el-class-10-maths-CBSE

Immediate cause of variations of A Mutations B Environmental class 10 biology CBSE

A rectangular container whose base is a square of side class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Why is Sardar Vallabhbhai Patel called the Iron man class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Write an application to the principal requesting five class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

