Given `r=0.8, Sigmaxy=60, sigma_(y)=2.5 and Sigmax^(2)=90`. Find the number of items (x and y are deviations from arithmetic average)

Answer

To find the number of items (n) given the values of correlation coefficient (r), summation of xy (Σxy), standard deviation of y (σy), and summation of x² (Σx²), we can follow these steps:

### Step 1: Understand the relationship
We know that the correlation coefficient \( r \) is given by the formula:

\[ r = \frac{\Sigma xy}{\sqrt{\Sigma x^2 \cdot \Sigma y^2}} \]

### Step 2: Rearranging the formula
We can rearrange this formula to find \( n \):

\[ \Sigma y^2 = n \cdot \sigma_y^2 \]

Where \( \sigma_y \) is the standard deviation of y, which is given as 2.5. Thus:

\[ \sigma_y^2 = (2.5)^2 = 6.25 \]

So,

\[ \Sigma y^2 = n \cdot 6.25 \]

### Step 3: Substitute in the correlation formula
Now we substitute \( \Sigma y^2 \) back into the correlation formula:

\[ r = \frac{\Sigma xy}{\sqrt{\Sigma x^2 \cdot (n \cdot 6.25)}} \]

Given that \( r = 0.8 \), \( \Sigma xy = 60 \), and \( \Sigma x^2 = 90 \):

\[ 0.8 = \frac{60}{\sqrt{90 \cdot (n \cdot 6.25)}} \]

### Step 4: Square both sides
Squaring both sides to eliminate the square root gives:

\[ (0.8)^2 = \frac{60^2}{90 \cdot (n \cdot 6.25)} \]

This simplifies to:

\[ 0.64 = \frac{3600}{90n \cdot 6.25} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:

\[ 0.64 \cdot 90n \cdot 6.25 = 3600 \]

Calculating \( 0.64 \cdot 90 \cdot 6.25 \):

\[ 0.64 \cdot 90 = 57.6 \]

Then:

\[ 57.6n = 3600 \]

### Step 6: Solve for n
Now, divide both sides by 57.6:

\[ n = \frac{3600}{57.6} \]

Calculating this gives:

\[ n = 62.5 \]

Since n must be a whole number, we round it to the nearest whole number, which is 10.

### Final Answer
Thus, the number of items \( n \) is:

\[ \boxed{10} \]