Question

Q6 The weight of adults in USA is normally distributed with a mean of 172 pounds and a standard deviation of 29 pounds. What is the probability that a single adult will weigh more than 190 pounds?

Q7 Along the lines of Q6 above, what is the probability that 25 randomly selected adults will have a MEAN more than 190 pounds?

Q8 Along the lines of Q6 above, an elevator has a sign that says that the maximum allowable weight is 4750 pounds. If 25 randomly selected people cram into the elevator, what is the probability that it will be over the maximum allowable weight?

Q9 The human gestation period (pregnancy period) is normally distributed with a mean of 268 days and a standard deviation of 15 days. If 25 women are randomly selected, find the probability that the sample will have a mean of less than 260 days.

Q10 Along the lines of Q9 above, a random selection of 25 woman (volunteers) are put on a special diet and the sample mean is less than 260 days. Does it appear that the diet has an effect of gestation period? What could make you more “certain”?

Answer 1

6)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	172
std deviation   =σ=	29.0000

probability =

P(X>190)

=

P(Z>0.62)=

1-P(Z<0.62)=

1-0.7324=

0.2676

7)

sample size       =n=	25
std error=σx̅=σ/√n=	5.8000

probability =

P(X>190)

=

P(Z>3.1)=

1-P(Z<3.1)=

1-0.9990=

0.0010

8)

probability that it will be over the maximum allowable weight =P(Xbar>4750/25)=P(Xbar>190)=0.0010

9)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	268
std deviation   =σ=	15.0000
sample size       =n=	25
std error=σx̅=σ/√n=	3.0000

probability =

P(X<260)

=

P(Z<-2.67)=

0.0038

10)

as probability of above event is less then 0.05 ; therefore this is an ususual event hence diet has an effect of gestation period