Question #d4932

Aug 26, 2017

See a solution process below:

Explanation:

The mean is calculated by adding a set of values and then dividing the sum by the number of values added. The formula is:

$A = \frac{s}{n}$

Where:

$A$ is the median or average. For the first 7 days in January we are told this is ${3}^{o} \text{C}$

$s$ is the sum of the numbers. What we will solve for in this part of the problem.

$n$ is the number of numbers. $7$ for this problem for the first $7$ days in January.

Substituting and solving for $s$ gives:

${3}^{o} \text{C} = \frac{s}{7}$

$\textcolor{red}{7} \times {3}^{o} \text{C} = \textcolor{red}{7} \times \frac{s}{7}$

${21}^{o} \text{C} = \cancel{\textcolor{red}{7}} \times \frac{s}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}}$

${21}^{o} \text{C} = s$

$s = {21}^{o} \text{C} +$

To determine the mean for the $8$ days we need to find the sum of the temperatures for the $8$ days and divide by the number of days, which will now be $8$.

To find the sum of the $8$ days, we take the sum of the first $7$ days which we just calculated and add the temperature from the $8$th day.

${s}_{8} = {21}^{o} \text{C" + 8^o"C" = 29^o"C}$

Substituting this new sum and dividing by $8$ days gives:

$A = \frac{{29}^{o} \text{C}}{8}$

$A = {3.625}^{o} \text{C}$

The mean temperature for the first 8 days in January is: ${3.625}^{o} \text{C}$

Aug 26, 2017

$3.625$°C

Explanation:

Let ${T}_{n}$ denote temperature on the nth day. Let ${M}_{n}$ denote the mean temperature over the first n days.
Mean for the first seven days is

${M}_{7} = \frac{{T}_{1} + {T}_{2} + \ldots + {T}_{7}}{7}$

$7 \cdot {M}_{7} = {T}_{1} + {T}_{2} + \ldots + {T}_{7}$

${M}_{8} = \frac{{T}_{1} + {T}_{2} + \ldots + {T}_{7} + {T}_{8}}{8}$

$\implies {M}_{8} = \frac{7 \cdot {M}_{7} + {T}_{8}}{8}$

$\therefore {M}_{8} = \frac{7 \cdot 3 + 8}{8} = \frac{29}{8} = 3.625$°C