Question

Lengths of human pregnancies are normally distributed with a mean of 280 days and a standard deviation of 15 days.   

A. What is the probability that a baby is born before 37 weeks?   

B. Between what two values lie the middle 95% of pregnancy lengths?

C. Between what two values lie the middle 90% of pregnancy lengths?


The first exam for Tadd’s sections of Stat 1040 has a mean of 75 and a stdev of 13.7.   

D. What score would you have to get in order to be in the top 20% of the class on exam

E. What is the probability that someone gets an A on the exam (a score of 90 or above)?

The variable Z has a standard normal distribution. (Round your answers to two decimal places.) (a) Find the number z that has cumulative proportion 0.63. (b) Find the number z such that the event Z > z has proportion 0.37

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	280
std deviation   =σ=	15.0000

probability that a baby is born before 37 weeks (259 days):

probability =

P(X<259)

=

P(Z<-1.4)=

0.0808

b)

for middle 95% values ; critical z =-/+1.96

therefore corresponding values =mean -/+ z*std deviation =280-/+1.96*15=250.6 to 309.4 days

c)

for middle 95% values ; critical z =-/+1.28

therefore corresponding values =mean -/+ z*std deviation =280-/+1.28*15=260.8 to 299.2 days

d)

for top 20% ; critical z =0.84

hence corresponding values =mean + z*std deviation =75+0.84*13.7=86.51

E)

probability =

P(X>90)

=

P(Z>1.09)=

1-P(Z<1.09)=

1-0.8621=

0.1379

2)

a)z =0.33

b)z=0.33