Question

The information below is the number of daily emergency service calls made by the volunteer ambulance service of Walterboro, South Carolina, for the last 50 days. To explain, there were 22 days on which there were two emergency calls, and 9 days on which there were three emergency calls.
Number of Calls Frequency


0 8


1 10


2 22


3 9


4 1


Total 50



a. Convert this information on the number of calls to a probability distribution. (Round your answers to 2 decimal places.)



Number of calls Probability



0



1



2



3



4



b. Is this an example of a discrete or continuous probability distribution?

Answer 1

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: The number of emergency service calls made by the volunteer ambulance service of Walterboro, South Carolina for 50 days.

a)

In the first column, you have the possible number of emergency calls made, in the second column, you have the observed absolute frequency (fi) for each value of the variable.

To establish the probability distribution you have to calculate the relative frequency (hi) for each value of X.

The formula for the relative frequency is

hi= fi*n where i= 0, 1, 2, 3, 4

X₁= 0

f₁= 8

h₁= 8/50= 0.16

X₂= 1

f₂= 10

h₂= 0.20

X₃= 2

f₃= 22

h₃= 22/50= 0.44

X₄= 3

f₄= 9

h₄= 9/50= 0.18

X₅= 4

f₅= 1

h₅= 1/50= 0.02

(See attachment)

b)

The variable of interest is a cuantitative discrete variable, so this is an example of a driscrete distribution.

Discrete variables are numerical and take certain numbers within its range of definition, contrary to the continuous variables that can take any value within the range of definition of the variable.

Another example of a discrete variable is "the money", If the variable is "I have at most 10 dollars in my wallet", the range of definition goes from 0 dollars to 10 dollars.

c) To calculate the mean when the data is arranged in a frequency table you have to use the following formula:

X[bar]= ∑Xihi= (0*0.16)+(1*0.2)+(2*0.44)+(3*0.18)+(4*0.02)= 1.7

This means that they expect an average of 1.7 calls per day.

d) The formula for the standard deviation is:

$S=\sqrt{\frac{1}{n-1}[sumX^2fi-\frac{(sumXfi)^2}{n} ] }$

∑X²fi= (0²*8)+(1²*10)+(2²*22)+(3²*9)+(4²*1)=  195

∑Xfi= (0*8)+(1*10)+(2*22)+(3*9)+(4*1)= 85

$S=\sqrt{\frac{1}{49}[195-\frac{(85)^2}{50} ] }= 1.015$

I hope this helps!