Question

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.

A) 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

Solution :

Given that ,

mean = = 60

standard deviation = = 4

a) P(x 54.2) = P[(x - ) / (54.2 - 60) / 4]

= P(z -1.45)

Using z table,

= 0.0735

b) P(x 54.2 ) = 1 - P(x 54.2)

= 1 - P[(x - ) / (54.2 - 60) / 4 ]

= 1 -  P(z -1.45)   

  Using z table,

= 1 - 0.0735

= 0.9265

c) P( 54.2 < x < 65.8) = P[(54.2 - 60)/ 4) < (x - ) /  < (65.8 - 60) / 4) ]

= P( -1.45 < z < 1.45)

= P(z < 1.45) - P(z < -1.45)

Using z table,

= 0.9265 - 0.0735

= 0.8530

d) n = 16

=   = 60

= / n = 4/ 16 = 1

P( > 60.85) = 1 - P( < 60.85)

= 1 - P[( - ) / < (60.85 - 60) / 1]

= 1 - P(z < 0.85 )

Using z table,    

= 1 - 0.8023

= 0.1977