Q6 The weight of adults in USA is normally distributed with a mean of 172 pounds...

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

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

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

Q9 The human gestation period (pregnancy period) is normallydistributed with a mean of 268 days and a standard deviation of 15days. If 25 women are randomly selected, find the probability thatthe 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 lessthan 260 days. Does it appear that the diet has an effect ofgestation period? What could make you more “certain”?

Answer & Explanation Solved by verified expert
3.6 Ratings (351 Votes)

6)

for normal distribution z score =(X-?)/?
here mean=       ?= 172
std deviation   =?= 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=?x?=?/?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-?)/?
here mean=       ?= 268
std deviation   =?= 15.0000
sample size       =n= 25
std error=?x?=?/?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


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