Question

In: Statistics and Probability

Each sweat shop worker at a computer factory can put together 4.2 computers per hour on...

Each sweat shop worker at a computer factory can put together 4.2 computers per hour on average with a standard deviation of 1 computers. 15 workers are randomly selected to work the next shift at the factory. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of X X ? X X ~ N(,) What is the distribution of ¯ x x¯ ? ¯ x x¯ ~ N(,) What is the distribution of ∑ x ∑x ? ∑ x ∑x ~ N(,) If one randomly selected worker is observed, find the probability that this worker will put together between 4.3 and 4.5 computers per hour. For the 15 workers, find the probability that their average number of computers put together per hour is between 4.3 and 4.5. Find the probability that a 15 person shift will put together between 61.5 and 66 computers per hour. For part e) and f), is the assumption of normal necessary? YesNo A sticker that says "Great Dedication" will be given to the groups of 15 workers who have the top 15% productivity. What is the least total number of computers produced by a group that receives a sticker? minutes (round to the nearest computer)

Solutions

Expert Solution

1)X ~ N(4.2,1)

2)x¯ ~ N(4.2,0.2582)

3)∑x ~ N (63 , 3.8730)

4)

If one randomly selected worker is observed, find the probability that this worker will put together between 4.3 and 4.5 computers per hour:

probability =P(4.3<X<4.5)=P((4.3-4.2)/1)<Z<(4.5-4.2)/1)=P(0.1<Z<0.3)=0.6179-0.5398=0.0781

5)

For the 15 workers, find the probability that their average number of computers put together per hour is between 4.3 and 4.5 :

probability =P(4.3<X<4.5)=P((4.3-4.2)/0.258)<Z<(4.5-4.2)/0.258)=P(0.39<Z<1.16)=0.8774-0.6507=0.2266

6)

Find the probability that a 15 person shift will put together between 61.5 and 66 computers per hour. :

probability =P(61.5<X<66)=P((61.5-63)/3.873)<Z<(66-63)/3.873)=P(-0.39<Z<0.77)=0.7807-0.3493=0.4314

7)

Yes

8)

for 85th percentile critical value of z= 1.04
therefore corresponding value=mean+z*std deviation= 67.0141

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