A population biologist is studying a certain species of lizard, whose sexes appe

Question

A population biologist is studying a certain species of lizard, whose sexes appear alike, except for size. It is known that in the adult male population, length M is normally distributed with mean mu_M = 10.0 cm and standard deviation sigma_M = 2.5 cm, while in the adult female population, length F is normally distributed with mu_F = 16.0 cm and standard deviation sigma_F = 5.0 cm. Suppose that a single adult specimen of length 11 cm is captured at random, and its sex identified as either a larger-than-average male, or a smaller-than-average female. Calculate the probability that a randomly selected adult male is as large as, or larger than, this specimen. Calculate the probability that a randomly selected adult female is as small as. or smaller than, this specimen. Based on this information, which of these two events is more likely?. Repeat part (a) for a second captured adult specimen, of length 12 cm. Repeat part (a) for a third captured adult specimen, of length 13 cm.

Explanation / Answer

a)

i.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    11      
u = mean =    10      
          
s = standard deviation =    2.5      
          
Thus,          
          
z = (x - u) / s =    0.4      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.4   ) =    0.344578258 [ANSWER]

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ii)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    11      
u = mean =    16      
          
s = standard deviation =    5      
          
Thus,          
          
z = (x - u) / s =    -1      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1   ) =    0.158655254 [ANSWER]

Hence, it is more likely that it is A LARGER THAN AVERAGE MALE. [ANSWER]

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b)

i.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    12      
u = mean =    10      
          
s = standard deviation =    2.5      
          
Thus,          
          
z = (x - u) / s =    0.8      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.8   ) =    0.211855399 [ANSWER]

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ii.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    12      
u = mean =    16      
          
s = standard deviation =    5      
          
Thus,          
          
z = (x - u) / s =    -0.8      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.8   ) =    0.211855399 [ANSWER]

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HENCE, EACH EVENT ARE AS LIKELY AS EACH OTHER. [ANSWER]

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Hi! Please submit the next part (part c) as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!

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