The weights of 1500 ball bearings are normally distributed with mean 22.40 and standard
deviation 0.048. If 300 random samples of size 36 each are drawn from this population,
determine the expected mean and standard deviation of the sampling distribution of
means, if Sampling is done with replacement.
How many of the random samples in the above problem would have their means between
22·39 and 22·41?
Answers
Answer:
Step-by-step explanation:
\huge \sf\underline\color{blue}{Given}\dag
Given†
If the mode of 5,3,3,2,2xis3.then find x
\huge \sf\underline\color{blue}{Solution}\dag
Solution†\huge \sf\underline\color{blue}{Given}\dag
Given†
If the mode of 5,3,3,2,2xis3.then find x
\huge \sf\underline\color{blue}{Solution}\dag
Solution†
Let’s represent the series in a different perspective,
Below are the positions,
Name Positions 1st 2nd 3rd
So, representation would be,
Positions
1 2 3 (Position's Alias/Naming)
1 5 8 = 76
2 7 3 = 25
3 4 9 = 89
4 5 7 = 69
5 3 8 = ?
Now, (by observation)
Take 1st and 3rd positions of each expression to make a number like this,
Number = 3rd position number (concatenate) 1st position number
e.g., if expression is 2 5 8 Then Number would be 82.
Then subtract 2nd position number from the Number would be 82 - 5 = 77.
Therefore,
By Observation,
5 3 8 = 85(Number) - 3 = 82
Answer = 82.
LETS TRY TO ANSWER THIS
FIRST OF ALL FIND THE LOGIC BEHIND IT
IF THE NUMBER ARE
Let a,b,c = d
Then c×10 -(b-a)=d
so now going by this logic we have
1,5,8=76
8×10-(5–1) =76
2,7,3=25
3×10 -(7–2)=25
3,4,9=89
\huge \sf\underline\color{blue}{Given}\dag
Given†
If the mode of 5,3,3,2,2xis3.then find x
\huge \sf\underline\color{blue}{Solution}\dag
Solution†
Let’s represent the series in a different perspective,
Below are the positions,
Name Positions 1st 2nd 3rd
So, representation would be,
Positions
1 2 3 (Position's Alias/Naming)
1 5 8 = 76
2 7 3 = 25
3 4 9 = 89
4 5 7 = 69
5 3 8 = ?
Now, (by observation)
Take 1st and 3rd positions of each expression to make a number like this,
Number = 3rd position number (concatenate) 1st position number
e.g., if expression is 2 5 8 Then Number would be 82.
Then subtract 2nd position number from the Number would be 82 - 5 = 77.
Therefore,
By Observation,
5 3 8 = 85(Number) - 3 = 82
Answer = 82.
LETS TRY TO ANSWER THIS
FIRST OF ALL FIND THE LOGIC BEHIND IT
IF THE NUMBER ARE
Let a,b,c = d
Then c×10 -(b-a)=d
so now going by this logic we have
1,5,8=76
8×10-(5–1) =76
2,7,3=25
3×10 -(7–2)=25
3,4,9=89
9×10-(4–3)=89
4,5,7=69
7×10-(5–4)=69
So, 5,3,8=????
8×10-(3–5)=82
The answer is 82
9×10-(4–3)=89
4,5,7=69
7×10-(5–4)=69
So, 5,3,8=????
8×10-(3–5)=82
The answer is 82
Let’s represent the series in a different perspective,
Below are the positions,
Name Positions 1st 2nd 3rd
So, representation would be,
Positions
1 2 3 (Position's Alias/Naming)
1 5 8 = 76
2 7 3 = 25
3 4 9 = 89
4 5 7 = 69
5 3 8 = ?
Now, (by observation)
Take 1st and 3rd positions of each expression to make a number like this,
Number = 3rd position number (concatenate) 1st position number
e.g., if expression is 2 5 8 Then Number would be 82.
Then subtract 2nd position number from the Number would be 82 - 5 = 77.
Therefore,
By Observation,
5 3 8 = 85(Number) - 3 = 82
Answer = 82.
LETS TRY TO ANSWER THIS
FIRST OF ALL FIND THE LOGIC BEHIND IT
IF THE NUMBER ARE
Let a,b,c = d\huge \sf\underline\color{blue}{Given}\dag
Given†
If the mode of 5,3,3,2,2xis3.then find x
\huge \sf\underline\color{blue}{Solution}\dag
Solution†
Let’s represent the series in a different perspective,
Below are the positions,
Name Positions 1st 2nd 3rd
So, representation would be,
Positions
1 2 3 (Position's Alias/Naming)
1 5 8 = 76
2 7 3 = 25
3 4 9 = 89
4 5 7 = 69
5 3 8 = ?
Now, (by observation)
Take 1st and 3rd positions of each expression to make a number like this,
Number = 3rd position number (concatenate) 1st position number
e.g., if expression is 2 5 8 Then Number would be 82.
Then subtract 2nd position number from the Number would be 82 - 5 = 77.
Therefore,
By Observation,
5 3 8 = 85(Number) - 3 = 82
Answer = 82.
LETS TRY TO ANSWER THIS
FIRST OF ALL FIND THE LOGIC BEHIND IT
IF THE NUMBER ARE
Let a,b,c = d
Then c×10 -(b-a)=d
so now going by this logic we have
1,5,8=76
8×10-(5–1) =76
2,7,3=25
3×10 -(7–2)=25
3,4,9=89
9×10-(4–3)=89
4,5,7=69
7×10-(5–4)=69
\huge \sf\underline\color{blue}{Given}\dag
Given†
If the mode of 5,3,3,2,2xis3.then find x
\huge \sf\underline\color{blue}{Solution}\dag
Solution†
Let’s represent the series in a different perspective,
Below are the positions,
Name Positions 1st 2nd 3rd
So, representation would be,
Positions
1 2 3 (Position's Alias/Naming)
1 5 8 = 76
2 7 3 = 25
3 4 9 = 89
4 5 7 = 69
5 3 8 = ?
Now, (by observation)
Take 1st and 3rd positions of each expression to make a number like this,
Number = 3rd position number (concatenate) 1st position number
e.g., if expression is 2 5 8 Then Number would be 82.
Then subtract 2nd position number from the Number would be 82 - 5 = 77.
Therefore,
By Observation,
5 3 8 = 85(Number) - 3 = 82
Answer = 82.
LETS TRY TO ANSWER THIS
FIRST OF ALL FIND THE LOGIC B
The answer is 82
So, 5,3,8=????
8×10-(3–5)=82
The answer is 82
So, 5,3,8=????
8×10-(3–5)=82
The answer is 82
Answer:
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