Question

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 sample variance and standard deviation for this data set. (Round your answers to four decimal places.) Variance Standard deviation

Answer 1

Answer:

$Mean = 48.6364$

$s^2 = 331.8545$ -- Sample Variance

$s = 18.2169$ --- Sample Standard Deviation

Step-by-step explanation:

Given

$n = 11$

$Data: 24, 41 ,25 ,68 ,53 ,83 ,43 ,63 ,56 ,34 ,45$

Solving (a): The sample mean

$Mean = \frac{\sum x}{n}$

So:

$Mean = \frac{24 +41 +25 +68 +53 +83 +43 +63 +56 +34+ 45}{11}$

$Mean = \frac{535}{11}$

$Mean = 48.6364$

Solving (b): Sample variance (s^2)

This is calculated as:

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

Where:

$\bar x = 48.6364$

So:

$s^2 = \frac{(24 -48.6364)^2 +(41 -48.6364)^2 +......................+( 45 -48.6364)^2}{11-1}$

$s^2 = \frac{3318.54545456}{10}$

$s^2 = 331.8545$

Solving (c): Sample Standard Deviation (s)

This is calculated as:

$s = \sqrt {s^2$

$s = \sqrt{331.8545$

$s = 18.2169$