Question

1. using the following informal histogram, calculate each of the following.

a. ( n )
b. ( sigma x )
c. ( sigma x^2 )

Explanation:

Part a: Calculate \( n \)

Step 1: Determine frequencies for each \( X \)

For \( X = 1 \), frequency \( f = 1 \); \( X = 2 \), \( f = 3 \); \( X = 3 \), \( f = 3 \); \( X = 4 \), \( f = 4 \); \( X = 5 \), \( f = 2 \); \( X = 6 \), \( f = 1 \).

Step 2: Sum the frequencies

\( n=\sum f = 1 + 3 + 3 + 4 + 2 + 1 \)
\( n = 14 \)

Part b: Calculate \( \sum X \)

Step 1: Multiply each \( X \) by its frequency

\( 1\times1 = 1 \), \( 2\times3 = 6 \), \( 3\times3 = 9 \), \( 4\times4 = 16 \), \( 5\times2 = 10 \), \( 6\times1 = 6 \)

Step 2: Sum these products

\( \sum X=1 + 6 + 9 + 16 + 10 + 6 \)
\( \sum X = 48 \)

Part c: Calculate \( \sum X^{2} \)

Step 1: Square each \( X \) and multiply by its frequency

\( 1^{2}\times1 = 1 \), \( 2^{2}\times3 = 12 \), \( 3^{2}\times3 = 27 \), \( 4^{2}\times4 = 64 \), \( 5^{2}\times2 = 50 \), \( 6^{2}\times1 = 36 \)

Step 2: Sum these products

\( \sum X^{2}=1 + 12 + 27 + 64 + 50 + 36 \)
\( \sum X^{2}=190 \)

Answer:

s:
a. \( \boldsymbol{n = 14} \)
b. \( \boldsymbol{\sum X = 48} \)
c. \( \boldsymbol{\sum X^{2}=190} \)