Question

A sample of 26women is​ obtained, and their heights​ (in inches) and pulse rates​ (in beats per​ minute) are measured. The linear correlation coefficient is 0.208and the equation of the regression line is ModifyingAbove y with caret equals 17.8 plus 0.940 x​,where x represents height. The mean of the 26heights is 63.5in and the mean of the 26pulse rates is 77.9beats per minute. Find the best predicted pulse rate of a woman who is 71in tall.

Answer 1

Answer:

The pulse rate is 84.54bpm

Step-by-step explanation:

Given

$\^{} y = 17.8 + 0.940x$

Required

Find $\^{}y$ when $x = 71$

Substitute 71 for x in: $\^{} y = 17.8 + 0.940x$

$\^{} y = 17.8 + 0.940 * 71$

$\^{} y = 17.8 + 66.74$

$\^{} y = 84.54$

The pulse rate is 84.54bpm