Suppose the lengths of the pregnancies of a certain animal are approximately nor

Question

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean H - 124 days and standard deviation o 6 days. Complete parts (a) through (c). What is the probability that a randomly selected pregnancy lasts less than 122 days? The probability that a randomly selected pregnancy lasts less than 122 days is approximately [L]. (Round to four decimal places as needed.) What is the probability that a random sample of 26 pregnancies has a mean gestation period of less than 122 days? The probability that the mean of a random sample of 26 pregnancies is less than 122 days is approximately | |. (Round to four decimal places as needed.) What is the probability that a random sample of 53 pregnancies has a mean gestation period of less than 122 days? The probability that the mean of a random sample of 53 pregnancies is less than 122 days is approximately Q. (Round to four decimal places as needed.)

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    122      
u = mean =    124      
          
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) / s =    -0.333333333      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.333333333   ) =    0.36944134 [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 =    122      
u = mean =    124      
n = sample size =    26      
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.699673171      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.699673171   ) =    0.044596209 [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    122      
u = mean =    124      
n = sample size =    53      
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.426703296      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.426703296   ) =    0.007618356 [ANSWER]

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