Use the normal model N 110158 for the weights of the steers

Use the normal model N (1101,58) for the weights of the steers.

A) What weight represents the 57th quartile?

B) What weight represents the 90th quartile?

C) Whats the IQR of the weights of these steers?

Solution

Do you mean percentile instead of quartile? I assumed so, as we only have 4 quartiles.

I assume that 58 here is the variance, not the standard deviation.

So I use

s = sqrt(58) = 7.615773106

a)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.57      
          
Then, using table or technology,          
          
z =    0.176374165      
          
As x = u + z * s,          
          
where          
          
u = mean =    1101      
z = the critical z score =    0.176374165      
s = standard deviation =    7.615773106      
          
Then          
          
x = critical value =    1102.343226   [answer]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    1101      
z = the critical z score =    1.281551566      
s = standard deviation =    7.615773106      
          
Then          
          
x = critical value =    1110.760006      
  

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C)

As

IQR = 75th percentile - 25th percentile

for 75th percentile:

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,          
          
where          
          
u = mean =    1101      
z = the critical z score =    0.67448975      
s = standard deviation =    7.615773106      
          
Then          
          
x = critical value =    1106.136761      

For 25th percentile:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.25      
          
Then, using table or technology,          
          
z =    -0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    1101      
z = the critical z score =    -0.67448975      
s = standard deviation =    7.615773106      
          
Then          
          
x = critical value =    1095.863239      

Then,

IQr = Q3-Q1 = 1106.136761 - 1095.863239

IQR = 10.273522 [ANSWER]