15. What would happen if the genes involved in differentiation were
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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.
