Please explain and show each step The distribution of weight

Please explain and show each step.

The distribution of weights of fully-grown male Chinese alligators is approximately normal with a mean of 32 kg and a standard deviation of 6 kg. a. Which two weights are the first and third quartiles of this distribution? b. If you select a random group of 5 alligators, what\'s the probability that all of them weigh between 30 and 40 kg?

Solution

a)

FIRST QUARTILE:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.25 (first quartile)
          
Then, using table or technology,          
          
z =    -0.67448975      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    32      
z = the critical z score =    -0.67448975      
s = standard deviation =    6      
          
Then          
          
x = critical value =    27.9530615 [ANSWER, FIRST QUARTILE]

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THIRD QUARTILE:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.75      
          
Then, using table or technology,          
          
z =    0.67448975      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    32      
z = the critical z score =    0.67448975      
s = standard deviation =    6      
          
Then          
          
x = critical value =    36.0469385   [ANSWER, THIRD QUARTILE]

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B)

We first get the probability that an alligator weights 30 to 40 kg.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    30      
x2 = upper bound =    40      
u = mean =    32      
          
s = standard deviation =    6      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.333333333      
z2 = upper z score = (x2 - u) / s =    1.333333333      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.36944134      
P(z < z2) =    0.90878878      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.53934744      

Thus, the probability that all 5 will weigh between 30 and 40 is

P(all 5) = 0.53934744^5

= 0.045639734 [ANSWER, PART B]