Given: The systolic arterial blood pressure observed for 20 dogs is normally distributed with a mean of 152 mm of mercury (Hg) and a standard deviation of 18 mm of Hg.
To find: P(100 < 152)
Method: Calculation of Z-Score followed by the probability or area of the bell curve at X = 100.
Solution:
Mean u = 152, std s = 18
Z score = 
The value of P(100<152) is calculated by looking at the value of Z in the Z score for the standard normal distribution given in the image.
P(Z=-2.89) = 0.0019
The P(Z = -2.89) corresponds to the area in the left tail of the bell curve.
Thus the probability of 100 mm Hg blood pressure is 0.0019.
Animals- that have thick layer of blubber or fur
is the correct answer......
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Glucuronidation, a conjugation reaction, is thought to protect the liver by both reducing hepatic BA toxicity and increasing their urinary elimination. The present study evaluates the contribution of each process in the overall BA detoxification by glucuronidation.
Answer:
Carbon dioxide is the waste released