The mean incubation time of fertilized eggs is 21 days Suppo

The mean incubation time of fertilized eggs is 21 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day.

(a) Determine the 20th percentile for incubation times.

(b) Determine the incubation times that make up the middle 39% of fertilized eggs.

Solution

suppose x is a random variable that denotes the incubation times.

we are gven that x follows normal distribution with mean 21 and standard deviation is 1

a) we need to find x1 (20th percentile) such that

p(x<x1) =.20   

p(z<z1) = .20

where

z= (x-21)/1 is a standard normal variate

and z1 = (x1-21)/1

now from standard normal table and p(z<z1) = .20

z1 = -0.8416212

(x1-21)/1 =  -0.8416212

therfore x1= 21 -0.8416212

= 21.84162 days (20th percentile of incubation time.

b)  we need to find the incubation times (x1 and x2) that make up the middle 39% of fertilized eggs.

p(x1<X<x2) = .39

p(X<x2)- p(X<x1)=.39

as we know that in normal dsitribution p(X>x2) = P(X<x1) and p(x<x1)=1-p(X<x2)

thus p(X<x2)- (1-p(X<x2))=.39 [by putting p(x<x1)=1-p(X<x2) ]

2*p(x<x2) -1 =.39

p(x<x2)= (1.39)/2

p(x<x2)= 0.695

p((x-21)/1 < (x2-21)/1) = 0.695 (converting normal variate to standard variate)

p(Z < (x2-21)/1) = 0.695

From standard normal table

(x2-21)/1 = 0.5100735

thus x2= 21.5100735

also we know that p(x<x1)=1-p(X<x2)

thus p(x<x1)= 1-0.695

p(X<x1)= 0.305

p((X-21)/1 < (x1-21)/1) = 0.305

(x1-21)/1 = -0.5100735 (from normal standard table)

x1= 21-0.5100735

x1=20.48993

thus 39% egs can be fertilized iff incubation time lies between 20.48993 and 21.51007