Answer:
P(X<240)=0.0521
P(240<X<270)=0.5466
280 days
Step-by-step explanation:
we are given the following information
mean=μ=266 days
standard deviation=σ=16 days
P(X<240)=P(X-μ/σ<240-266/16)=P(z<-1.625)
This probability can either be determined by utilizing normal distribution table or using Excel.
Using Excel function <em>NORM.S.DIST(-1.625,TRUE)</em> we get
P(X<240)=0.0521
P(240<X<270)=P((240-266)/16<X-μ/σ<(270-266)/16)=P(-1.625<z<0.25)
P(-1.625<z<0.25)=P(z<0.25)-P(z<-1625)
Using Excel function
<em>NORM.DIST(270,266,16,TRUE)-NORM.DIST(240,266,16,TRUE)</em> we get
P(240<X<270)=0.5466
P(X>x)=0.20
P(z>(x-266)/16)=0.20
Using Excel function <em>NORM.INV(0.8,266,16)</em> we get
x=280
<em>OR</em>
P(X>x)=0.2
The area to the right of x is 0.2 and the area between mean and x is
0.5-0.2=0.3
Looking for 0.3 in the normal distribution table we don't find 0.3 but the closest value is .3023 and it corresponds to z-value 0.85.
z=0.85
z=x-μ/σ
σ*z+μ=x
x=16*0.85+266
x=279.6≅280
The longest 20% of pregnancies lasts 280 days
