# (a) Interpretation: The table for the analysis of the six bottles of wine of same variety for residue sugar is given below- Bottle Percent (w/v) Residual sugar 1 0.99,0.84,1.02 2 1.02,1.13,1.17,1.02 3 1.25,1.32,1.13,1.20,1.12 4 0.72,0.77,0.61,0.58 5 0.90,0.92,0.73 6 0.70,0.88,0.72,0.73 The value of standard deviation for each set is to be determined. Concept introduction: The formula for the calculation of mean and standard deviation can be calculated as follows: x ¯ = x 1 + x 2 + ......... + x n N x ¯ = ∑ i = 1 N x i N s = ∑ i = 1 N x i 2 − ( ∑ i = 1 N x i ) 2 N N − 1

### Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

### Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

#### Solutions

Chapter A1, Problem A1.9QAP
Interpretation Introduction

## (a)Interpretation:The table for the analysis of the six bottles of wine of same variety for residue sugar is given below- Bottle Percent (w/v) Residual sugar 1 0.99,0.84,1.02 2 1.02,1.13,1.17,1.02 3 1.25,1.32,1.13,1.20,1.12 4 0.72,0.77,0.61,0.58 5 0.90,0.92,0.73 6 0.70,0.88,0.72,0.73 The value of standard deviation for each set is to be determined.Concept introduction:The formula for the calculation of mean and standard deviation can be calculated as follows:x¯=x1+x2+.........+xnNx¯=∑i=1NxiNs=∑i=1Nxi2−(∑i=1Nxi)2NN−1

Interpretation Introduction

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