# Although bats are not known for their eyesight, they are able to locate prey (mainly insects) by emitting high-pitched sounds and listening for echoes. A paper gave the following distances (in centimeters) at which a bat first detected a nearby insect. 24 41 25 68 53 83 43 63 56 34 45 A button hyperlink to the SALT program that reads: Use SALT. (a) Calculate the sample mean distance at which the bat first detects an insect. (Round your answer to four decimal places.) 48.6364 cm (b) Calculate the

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Answer:[tex]Mean = 48.6364[/tex][tex]s^2 = 331.8545[/tex] -- Sample Variance[tex]s = 18.2169[/tex] --- Sample Standard DeviationStep-by-step explanation:Given[tex]n = 11[/tex][tex]Data: 24, 41 ,25 ,68 ,53 ,83 ,43 ,63 ,56 ,34 ,45[/tex]Solving (a): The sample mean[tex]Mean = \frac{\sum x}{n}[/tex]So:[tex]Mean = \frac{24 +41 +25 +68 +53 +83 +43 +63 +56 +34+ 45}{11}[/tex][tex]Mean = \frac{535}{11}[/tex][tex]Mean = 48.6364[/tex]Solving (b): Sample variance (s^2)This is calculated as:[tex]s^2 = \frac{\sum (x - \bar x_i)^2}{n - 1}[/tex]Where:[tex]\bar x = 48.6364[/tex]So:[tex]s^2 = \frac{(24 -48.6364)^2 +(41 -48.6364)^2 +......................+( 45 -48.6364)^2}{11-1}[/tex][tex]s^2 = \frac{3318.54545456}{10}[/tex][tex]s^2 = 331.8545[/tex]Solving (c): Sample Standard Deviation (s)This is calculated as:[tex]s = \sqrt {s^2[/tex][tex]s = \sqrt{331.8545[/tex][tex]s = 18.2169[/tex]