In: Math
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. 23 40 27 56 52 34 42 61 68 45 83 (a) Compute the sample mean distance at which the bat first detects an insect. (Round your answer to three decimal places.) cm (b) Compute the sample variance and standard deviation for this data set. (Round your answers to two decimal places.) Variance?
Standard deviation?
Solution:
Given that,
23 40 27 56 52 34 42 61 68 45 83
n = 11
a )The mean of sample is
x/n = ( 23 + 40 + 27 + 56 + 52 + 34 + 42 + 61 + 68 + 45 + 83 / 11)
= 531 / 11
= 48.273
The sample mean is 48.273
Sample variance is s2
s2 = 1/(n-1)(x - )2
= 1/11 - 1 (23 - 48.273 )2+ (40 - 48.273)2+ (27 - 48.273 )2+ (56 - 48.273 )2 +( 52 - 48.273)2 + ( 34 - 48.273)2 +(42 - 48.273)2 + (61 - 48.273)2 + (68 - 48.273)2 + (45 - 48.273)2 + (83 - 48.273)2
= 1/10 (638.724529 + 68.442529 + 452.540529 + 59.706529 + 13.890529 + 203.718529 + 39.350529 + 161.976529 + 389.154529 + 10.712529 + 1205.964529)
= 3244.181819/10
= 324.42
sample variance = s2 = 324.42
The sample standard deviation is = s = s2 = 324.42 = 18.01