Solve the 3 following questions, please. The average weight of a box of cereal i

Question

Solve the 3 following questions, please.

The average weight of a box of cereal is 25.2 ounces with a standard deviation of 0.18 ounces. Quality control wants to make sure that the lightest 15% of cereal boxes are not sent to stores to sell. What would be the cutoff weight for these cereal boxes assuming the weight of boxes is normally distributed? Fifty numbers are rounded to the nearest integer and then summed. If the individual round-off-errors are uniformly distributed between -0.5 and 0.5, what is the approximate probability that the resultant sum differs from the exact sum by more than 3? An employee is selected from a staff of 10 to supervise a particular project by selecting a tag at random from a box containing 10 tags numbered 1 to 10. X represents the number on the tag that is drawn. (a) If a randomly drawn tag is drawn what is the probability that it was at least a 6? (b) If this process is repeated 50 times, with replacement, what is the probability that the average of all the tags drawn was at least a 6?

Explanation / Answer

23) for 15 percentile ; z=-1.0364

hence cutoff weight =25.2-0.18*1.0364=25.01

24) here mean error for one number =(-0.5+0.5)/2=0

std deviation =(0.5-(-0.5)/(12)1/2=0.2887

hence mean for 50 numbers =0*50=0

and std deviation =0.2887*(50)1/2=2.04

hence P(X>3)=1-P(X<3)=1-P(Z<(3-0)/2.04)=1-P(Z<1.4697)=1-0.9292=0.0708

25)

a) probabilty =5/10 =0.5 (as there are 5 number eequal or above 6.

b)mean of single observation =(1+2+3+4+...+10)/10=5.5

and std deviaiton=2.872

hence mean for 50 draws=5.5

and std deviation =2.872/(50)1/2 =0.4062

hence P(X>6)=1-P(X<6)=1-P(Z<(6-5.5)/0.4062)=1-P(Z<1.2309)=1-0.8908=0.1092

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