A bakery owner tabulated sales to 50 customers and obtained the distribution. Find the mean and standard deviation for the data. Sales in dollarsNumber of customers0.00−1,99182.00−3.99164.00−5.9996.00−7.9948.00−9.993

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Answered question

2021-12-05

A bakery owner tabulated sales to 50 customers and obtained the distribution. Find the mean and standard deviation for the data.

\begin{array}{cc} Sales in dollars & Number of customers \\ 0.00 - 1, 99 & 18 \\ 2.00 - 3.99 & 16 \\ 4.00 - 5.99 & 9 \\ 6.00 - 7.99 & 4 \\ 8.00 - 9.99 & 3 \end{array}

Answer & Explanation

Louis Gregory

Beginner2021-12-06Added 14 answers

Step 1
The table is given as:

\begin{array}{cc} Class & Continuous class & Class mark (x_{i}) & F r e q u e n c y (f_{i}) & f_{i} x_{i} & f_{i} x_{i}^{2} \\ 0.00 - 1, 99 & 0.00 - 1.995 & 1 & 18 & 18 & 18 \\ 2.00 - 3.99 & 1.995 - 3.995 & 3 & 16 & 48 & 144 \\ 4.00 - 5.99 & 3.995 - 5.995 & 5 & 9 & 45 & 225 \\ 6.00 - 7.99 & 5.995 - 7.995 & 7 & 4 & 28 & 196 \\ 8.00 - 9.99 & 7.995 - 9.995 & 9 & 3 & 27 & 243 \\ 50 & 166 & 826 \end{array}

Step 2
The mean is given by:

μ = \frac{\sum f_{i} x_{i}}{\sum f_{i}}

= \frac{166}{50}

= 3.32

The variance is given by:

V a r (X) = \frac{\sum f_{i} x_{i}^{2}}{\sum f_{i}} - μ^{2}

= \frac{826}{50} - {(3.32)}^{2}

= 5.4976

Step 3
The standard deviation is given by:

σ = \sqrt{V a r (X)}

= \sqrt{5.496}

\approx 2.3444

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