Question

coach ferguson uses a thermometer to measure the temperature (in degrees fahrenheit) at 20 different locations in the school swimming pool. an analysis of the data yields a median of 77°f and an interquartile range of 5°f. recall that °c = \\( \frac { 5 } { 9 } ( ^ { circ } f - 32 ) \\). (a) find the median temperature reading in degrees celsius. °c (do not round.) (b) calculate the interquartile range of the temperature readings in degrees celsius. °c (round to 2 decimal places.)

Explanation:

Step1: Convert median temperature from Fahrenheit to Celsius

Given the formula \(^{\circ}C=\frac{5}{9}(^{\circ}F - 32)\). For the median \(^{\circ}F = 77\), substitute into the formula:
\(^{\circ}C=\frac{5}{9}(77 - 32)\)
\(=\frac{5}{9}\times45\)
\(=25\)

Step2: Convert inter - quartile range from Fahrenheit to Celsius

The inter - quartile range (IQR) in Fahrenheit is \(IQR_{F}=5\).
Since \(y=\frac{5}{9}(x - 32)\), if \(x_1\) and \(x_2\) are two values, \(y_1=\frac{5}{9}(x_1 - 32)\) and \(y_2=\frac{5}{9}(x_2 - 32)\). Then \(y_2 - y_1=\frac{5}{9}(x_2 - x_1)\) (because the \(-32\) terms cancel out when subtracting).
So \(IQR_{C}=\frac{5}{9}\times IQR_{F}\)
Substitute \(IQR_{F} = 5\) into the formula: \(IQR_{C}=\frac{5}{9}\times5=\frac{25}{9}\approx2.78\)

Answer:

(a) \(25\)
(b) \(2.78\)