What sample size is need to construct a 98% confidence interval for the average

Question

What sample size is need to construct a 98% confidence interval for the average height u, of all 7 th graders if past studies have shown a standard deviation of 2.6 inches and a margin of error of 1 What sample size is need to construct a 98% confidence interval for the average height u, of all 7 th graders if past studies have shown a standard deviation of 2.6 inches and a margin of error of 1 3) The electronics store "TV's and More" wants to know how many large screen television sets to order for the next monthly delivery. Past data formed a normal distribution and shows that at this time of the year, they sell an average of 36 large screen television sets per month order 40 sets, what is the probability that they will run out of stock? , with a standard deviation of 8. If they The waiting time for customers at a popular restaurant before being seated for lunch are normally distributed with a mean of 16 minutes and a standard deviation of 4 minutes. Find the percent of customers who wait between 12 and 24 minutes before being seated. 4) If you are an engineer that designs motorcyele helmets fo the breadths of a man's head. After considerable research you found men have head breadths that are normally distributed with a mean of 5.9 inches and a ) r men you must consider standard deviation of 1.2 inches. This morning you are told about financial constraints and you must design the helmets to fit all men except those that have head breadths that are in the smallest or largest 2.8% of the population. What are the minimum and maximum head breadths that the helmets will fit?

Explanation / Answer

3) probability that will out of stock =P(demand is greater than 40 =P(X>40)=P(Z>(40-36)/8)=P(Z>0.5)

=1-P(Z<0.5)=1-0.6915 =0.3085

4)

P(12<X<24)=P((12-16)/4<Z<(24-16)/4)=P(-1<Z<2)=0.9772-0.1587 =0.8186

5)

for bottom 2.8% percentile z score =-1.911

hence corresponding mimum head breadth =mean +z*std deviation =5.9-1.911*1.2=3.61 inches

for top 2.8% percentile z score =1.911

hence corresponding maximum head breadth =mean +z*std deviation =5.9+1.911*1.2=8.19

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question on top :

here for 98% CI ; critical value z=2.326

and std deviaiton =2.6

margin of error E =1

hence required sample size n=(z*std deviation/E)2 =(2.326*2.6/1)2 =~37

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