Question

The lengths of earthworms are normally distributed with a mean of 3.2 inches

and a standard deviation of 0.8 inches.

A. What is the probability that a sample of 40 worms has a mean length of at least 3 inches?

B. What is the probability that Freddy the earthworm is between 2.5 in and 4.0 in long?

C. If I dig up a random sample of 25 earthworms, what is the probability that the mean nl _______

length in the sample would be less than 3.5 inches?

Answer 1

a)

Solution :

Given that,

mean = = 3.2

standard deviation = =0.8

n=40

= =3.2

= / n = 0.8 / 40 = 0.1265

P( > 3) = 1 - P( < 3)

= 1 - P[( - ) / < (3-3.2) / 0.1265]

= 1 - P(z < -1.58)

Using z table

= 1 - 0.0571

= 0.9429

probability= 0.9429

b)

Solution :

Given that ,

mean =   = 3.2

standard deviation = = 0.8  

P(2.5< x <4.0 ) = P[(2.5-3.2) /0.8 < (x - ) / < (4.0-3.2) / 0.8)]

= P(-0.88 < Z <1 )

= P(Z < 1) - P(Z <-0.88 )

Using z table   

= 0.8413 - 0.1894

probability= 0.6465

c)

Solution :

Given that ,

mean =   = 3.2

standard deviation = σ   = 0.8

n = 25

= 3.2

=  / n = 0.8 / 25=0.16

P( < 3.5) = P[( - ) / < (3.5-3.2) / 0.16]

= P(z <1.88 )

Using z table  

= 0.9699   

probability= 0.9699