Question

Question 6 options:

The length of western rattlesnakes are normally distributed with a mean of 60 inches and a standard deviation of 4 inches.

Enter answers as a decimal rounded to 4 decimal places with a 0 to the left of the decimal point.
Do not enter an answer as a percent.

Suppose a rattlesnake is found on a mountain trail:

a. What is the probability that the rattlesnakes' length will be equal to or less than 54.2 inches?

b. What is the probability its' length will be equal to or greater than 54.2 inches?

c. What is the probability that the rattlesnakes' length will be between 54.2 inches and 65.8 inches?

d. Suppose a nest of 16 rattlesnakes are found on the mountain trail:

What is the probability that the average length of the rattlesnakes will be 60.85 inches or more?

Answer 1

(a)

= 60

= 4

To find P(X54.2):
Z = (54.2 - 60)/4

= - 1.45

Table of Area Under Standard Normal Curve gives area = 0.4265

So,

P(X54.2) = 0.5 - 0.4265 = 0.0735

So,

Answer is:

0.0735

(b)

= 60

= 4

To find P(X54.2):
Z = (54.2 - 60)/4

= - 1.45

Table of Area Under Standard Normal Curve gives area = 0.4265

So,

P(X54.2) = 0.5 + 0.4265 = 0.9265

So,

Answer is:

0.9265

(c)

= 60

= 4

To find P(54.2<X<65.8):

Case 1: For X from 54.2 to mid value:
Z = (54.2 - 60)/4

= - 1.45

Table of Area Under Standard Normal Curve gives area = 0.4265

Case 2: For X from mid value to 65.8:
Z = (65.8 - 60)/4

= 1.45

Table of Area Under Standard Normal Curve gives area = 0.4265

So,

P(54.2<X<65.8) = 2 X 0.4265 = 0.8530

So,

Answer is:

0.8530

(d)

= 60

= 4

n = 16

SE = /

= 4/

= 1

To find P(60.85):
Z = (60.85 - 60)/1

= 0.85

Table of Area Under Standard Normal Curve gives area = 0.3023

So,

P(X54.2) = 0.5 + 0.3023 = 0.8023

So,

Answer is:

0.8023