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The following information provided on ultimate tensile strength for a sample:

Sample size, \(n = 4\)

Sample mean,\(\bar x = 76831\)

Sample standard deviation,\(s = 180\)

Smallest \({x_i}\)= 76,683

Largest \({x_i}\)= 77048

Let\({x_m}\)and\({x_n}\)be the middle two sample observations. The formula of finding sample mean given by,

\(\bar x = \frac{{\sum {{x_i}} }}{n}\)

The first and last observations are 76683 and 77048 respectively. The value of sample mean is\(\bar x = 76831\).

Substitute these values in the above formula to obtain the required sum of two observations.

\(\begin{aligned}76831 &= \frac{{\left( {76683 + {x_m} + {x_n} + 77048} \right)}}{4}\\76831 * 4 &= \left( {153731 + {x_m} + {x_n}} \right)\\307324 &= 153731 + {x_m} + {x_n}\\{x_m} + {x_n} &= 307324 - 153731\\\left( {{x_m} + {x_n}} \right) &= 153593\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \cdots (1)\end{aligned}\)

The formula of sample variance given by,

\({s^2} = \frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}\)

Substitute the values of sample standard deviation \(s = 180\)and sample size\(n = 4\) in the above formula,

\(\begin{aligned}{\left( {180} \right)^2} &= \frac{{\left( \begin{array}{l}{\left( {76683 - 76831} \right)^2} + {\left( {{x_m} - 76831} \right)^2}\\ + {\left( {{x_n} - 76831} \right)^2} + {\left( {77048 - 76831} \right)^2}\end{array} \right)}}{{4 - 1}}\\32400 &= \frac{{\left( {68993 + {{\left( {{x_m} - 76831} \right)}^2} + {{\left( {{x_n} - 76831} \right)}^2}} \right)}}{3}\\32400 * 3 &= \left( {68993 + {{\left( {{x_m} - 76831} \right)}^2} + {{\left( {{x_n} - 76831} \right)}^2}} \right)\\97200 - 68993 &= \left( {{{\left( {{x_m} - 76831} \right)}^2} + \left( {{x_n} - 76831} \right)} \right)\\28207 &= \left( {{{\left( {{x_m} - 76831} \right)}^2} + {{\left( {153593 - {x_m} - 76831} \right)}^2}} \right)\\28207 &= \left( {{{\left( {{x_m} - 76831} \right)}^2} + {{\left( {76762 - {x_m}} \right)}^2}} \right)\\{x_m} &= 76910\end{aligned}\)

Therefore, the first middle observation is 76910.

Substitute the value of \({x_m} = 76910\)in the equation (1) to obtain the value of second middle observation.

\(\begin{aligned}\left( {{x_m} + {x_n}} \right) &= 153593\\\left( {76910 + {x_n}} \right) &= 153593\\{x_n} &= 153593 - 76910\\ &= 76683\end{aligned}\)

Therefore, the second middle observation is 76683. Thus, the two middle values are 76910 and 76683.