Certainly! Let's break down the solution step-by-step.
### (a) Provide the formula to find the value of [tex]\( k \)[/tex] for the given situation.
Given:
- Mean [tex]\( \mu = 56 \)[/tex]
- Total number of data points [tex]\( N = 32 + k \)[/tex]
- Sum of frequencies times their values [tex]\( \Sigma fx = 2557 + 5k \)[/tex]
The mean formula for discrete data is given by:
[tex]\[ \text{Mean} (\mu) = \frac{\Sigma fx}{N} \][/tex]
Substitute the given values into the mean formula:
[tex]\[ 56 = \frac{2557 + 5k}{32 + k} \][/tex]
To find the equation in terms of [tex]\( k \)[/tex], cross-multiply:
[tex]\[ 56(32 + k) = 2557 + 5k \][/tex]
Simplify this:
[tex]\[ 1792 + 56k = 2557 + 5k \][/tex]
Rearrange the equation to solve for [tex]\( k \)[/tex]:
[tex]\[ 56k - 5k = 2557 - 1792 \][/tex]
[tex]\[ 51k = 765 \][/tex]
[tex]\[ k = \frac{765}{51} \][/tex]
### (b) Find the value of [tex]\( k \)[/tex].
From the simplified equation:
[tex]\[ k = \frac{765}{51} \][/tex]
[tex]\[ k = 15 \][/tex]
### (c) If [tex]\( k \)[/tex] increases by 5, calculate the new mean.
If [tex]\( k \)[/tex] increases by 5:
[tex]\[ k_{new} = 15 + 5 = 20 \][/tex]
The new total number of data points:
[tex]\[ N_{new} = 32 + k_{new} = 32 + 20 = 52 \][/tex]
The new sum of frequencies times their values:
[tex]\[ \Sigma fx_{new} = 2557 + 5 \times 20 = 2557 + 100 = 2657 \][/tex]
Calculate the new mean:
[tex]\[ \text{New Mean} = \frac{\Sigma fx_{new}}{N_{new}} = \frac{2657}{52} \][/tex]
### (d) If [tex]\( k \)[/tex] increases to 20, by how much does the new mean increase or decrease?
If [tex]\( k \)[/tex] increases to 20:
[tex]\[ k = 20 \][/tex]
The new total number of data points:
[tex]\[ N_{20} = 32 + 20 = 52 \][/tex]
The new sum of frequencies times their values:
[tex]\[ \Sigma fx_{20} = 2557 + 5 \times 20 = 2657 \][/tex]
Calculate the new mean:
[tex]\[ \text{New Mean}_{20} = \frac{2657}{52} \approx 51.09615384615385 \][/tex]
Calculate the difference between the new mean and the original mean:
[tex]\[ \text{Mean Difference} = \text{New Mean}_{20} - \text{Original Mean} = 51.09615384615385 - 56 = -4.903846153846153 \][/tex]
Thus, when [tex]\( k \)[/tex] increases to 20, the new mean decreases by approximately [tex]\( 4.903846153846153 \)[/tex].
