Isle royale, an island in lake superior, has provided an important study site of wolves and their prey. of special interest is the study of the number of moose killed by wolves. in the period from 1958 to 1974, there were 296 moose deaths identified as wolf kills. the age distribution of the kills is as follows. age of moose in yearsnumber killed by wolvescalf (0.5 yr)1-56-1011-1516-201074779603(a) for each age group, compute the probability that a moose in that age group is killed by a wolf. (round your answers to three decimal places.)0.5 1-5 6-10 11-15 16-20 (b) consider all ages in a class equal to the class midpoint. find the expected age of a moose killed by a wolf and the standard deviation of the ages. (round your answers to two decimal places.)μ= σ=

Part a) Probability that a moose in that age group is killed by a wolf

Part b)

         μ = 5.61 years

       σ = 4.97 years

Explanation:

1) Start by arranging the table to interpret the information:

Age of Moose in years          Number killed by woolves

Calf (0.5)                                                 107

1 - 5                                                           47

6 - 10                                                         79

11 - 15                                                        60

16 - 20                                                        3

You can  now verify the total number of moose deaths identified as wolf kills: 107 + 47 + 79 + 60 + 3 = 296, such as stated in the first part of the question.

2) First question: a) For each age, group, compute the probability that a moos in that age group is killed by a wolf.

i) Formula:

Probability = number of positive outcomes / total number of events.

ii) Probability that a moose in an age group is killed by a wolf = number of moose killed by a wolf in that age group / total number of moose deaths identified as wolf kills.

iii) P  (0.5 year) = 107 / 296 = 0.361

iv) P (1 - 5) = 47 / 296 = 0.159

v) P (6 - 10) = 79 / 296 = 0.267

vi) P (11 - 15) = 60 / 296 = 0.203

vii) P (16 - 20) = 3 / 296 = 0.010

3) Second question: b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages.

i) Class       midpoint

   0.5               0.5

   1 - 5           (1 + 5) / 2 = 3

   6 - 10         (6 + 10) / 2 = 8

   11 - 15         (11 + 15) / 2 = 13

   16 - 20       (16 + 20) = 18

ii) Expected age of a moose killed by a wolf = mean of the distribution = μ

μ = Sum of the products of each probability times its age (mid point)

μ = 0.5 ( 0.361) + 3 ( 0.159) + 8 ( 0.267) + 13 ( 0.203) + 18 ( 0.010) = 5.61 years

μ = 5.61 years ← answer

iii) Stantard deviation of the ages = σ

σ = square root of the variace

variance = s = sum of the products of the squares of the differences between the mean and the class midpoint time the probability.

s =  (0.5 - 5.61)² (0.361) + (3 - 5.61)² ( 0.159) + (8  - 5.61)² ( 0.267) + (13 - 5.61)² ( 0.203) + (18 - 5.61)² ( 0.010) = 24.65

σ = √ (24.65) = 4.97 years ← answer

step-by-step explanation:

once you graph it hit 2nd (the blue button) and graph

that takes you to a table of coordinates

on the left is x coordinates and on the right is y coordinates and an empty box

(you may have to scroll up/down in order to get ( x, 0)

d. millimeters per hour.

step-by-step explanation:

we are given that,

carlo used a rain gauge to measure the amount of water his lawn is receiving from the sprinklers.

since, according to the options, we see that,

water cannot be measured by 'miles per hour', 'meters per second' and 'grams per seconds'.

we get that, the water can be measured by 'millimeters per hour'.

hence, the appropriate rate for describing the amount of water is 'millimeters per hour'.