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elementary process The inside diameter of a randomly selected piston ring is a r

ID: 3154228 • Letter: E

Question

elementary process The inside diameter of a randomly selected piston ring is a random variable with mean value 11 cm and standard devation 0.03 cm. Suppose the distribution of the diameter is normal. (Round your answers to four decimal places.) Calculate P(10.99 le X le 11.01) when n = 16 P(10.99 le X le 11.01) = 3829 How likely is it that the sample mean diameter exceeds 11.01 when n = 25? P(X ge 11.01) = 2660 You may need to use the appropriate table in the Appendix of tables to answer this question. There are 46 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of 5 min and a standard deviation of 4 min. (Round your answers to four decimal places.) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins? If the sports report begins at 11:10, what is the probability that he misses part of the report if he waits until grading is done before turning on the TV? You may need to use the appropriate table in the Appendix of tables to answer this question.

Explanation / Answer

The inside diameter...

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a)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    10.99      
x2 = upper bound =    11.01      
u = mean =    11      
n = sample size =    16      
s = standard deviation =    0.03      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.333333333      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.333333333      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.09121122      
P(z < z2) =    0.90878878      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.817577561   [ANSWER]  

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    11.01      
u = mean =    11      
n = sample size =    25      
s = standard deviation =    0.03      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.666666667      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.666666667   ) =    0.047790352 [ANSWER]

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