12. The mean incubation time of fertilized eggs is 21 days. Suppose the incubati

Question

12. 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 17th percentile for incubation times.

(b) Determine the incubation times that make up the middle 95% of fertilized eggs.

(a) The 17th percentile for incubation times is _?__ days.

(Round to the nearest whole number as needed.)

(b) The incubation times that make up the middle 95% of fertilized eggs are ___?__ to __?__ days.

(Round to the nearest whole number as needed. Use ascending order.)

Explanation / Answer

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 (17th percentile) such that

p(x<x1) =.17  

p(z<z1) = .17

where

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

and z1 = (x1-21)/1

now from standard normal table and p(z<z1) = .17

z1 =-0.954165

(x1-21)/1 = -0.954165

therfore x1= 21 -0.954165

= 20.0458 days (17th percentile of incubation time.

b) we need to find the incubation times (x1 and x2) that make up the middle 95% of fertilized eggs.

p(x1<X<x2) = .95

p(X<x2)- p(X<x1)=.95

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))=.95 [by putting p(x<x1)=1-p(X<x2) ]

2*p(x<x2) -1 =.95

p(x<x2)= (1.95)/2

p(x<x2)= 0.975

p((x-21)/1 < (x2-21)/1) = 0.975 (converting normal variate to standard variate)

p(Z < (x2-21)/1) = 0.975

From standard normal table

(x2-21)/1 = 1.959964

thus x2= 22.959964

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

thus p(x<x1)= 1-0.975

p(X<x1)= 0.025

p((X-21)/1 < (x1-21)/1) = 0.025

(x1-21)/1 = -1.959964 (from normal standard table)

x1= 21-1.959964

x1=19.040036

thus 39% egs can be fertilized iff incubation time lies between 19.040036 and 22.959964

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