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There are two constants a and b.

\({y_i} = a{x_i} + b\)for i=1,2,鈥�,n.

The average temperature for initiating a certain chemical reaction is 87.3 degrees Celsius.

The sample standard deviation is 1.04.

The sample mean of x is given as,

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

The sample of y is given as,

\(\begin{array}{c}\bar y &=& \frac{{{y_1} + {y_2} + ... + {y_n}}}{n}\\ &=& \frac{{\left( {a{x_1} + b} \right) + \left( {a{x_2} + b} \right) + ... + \left( {a{x_n} + b} \right)}}{n}\\ &=& \frac{{a\left( {{x_1} + {x_2} + ... + {x_n}} \right) + nb}}{n}\\ &=& a\frac{{\sum {{x_i}} }}{n} + b\\ &=& a\bar x + b\end{array}\)

Thus, the relationship between \(\bar x\) and \(\bar y\) is \(\bar y = a\bar x + b\).

The sample variance of x is given as,

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

The sample variance of y is given as,

\(\begin{array}{c}s_y^2 &=& \frac{{\sum {{{\left( {{y_i} - \bar y} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{\sum {{{\left( {a{x_i} + b - \left( {a\bar x + b} \right)} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{\sum {{{\left( {a{x_i} + b - \left( {a\bar x + b} \right)} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{\sum {{{\left( {a{x_i} - a\bar x} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{{a^2}\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}\\ &=& {a^2}s_x^2\end{array}\)

Therefore, the relationship between \(s_x^2\) and \(s_y^2\) is: \(s_y^2 = {a^2}s_x^2\).

Fahrenheit in Celsius is given as,

\(F = \frac{9}{5}C + 32\)

Since the average is affected by the change of origin and scale.

The sample average measured in Fahrenheit is computed as,

\(\frac{9}{5}\left( {87.3} \right) + 32 = 189.14\)

Therefore, the sample average is 189.14 Fahrenheit.

Since the standard deviation is affected by only the change of scale.

The sample standard deviation measured in Fahrenheit is computed as,

\(\frac{9}{5}\left( {1.04} \right) = 1.872\)

Therefore, the sample standard deviation is 1.872 Fahrenheit.