Let x be a random variable representing the blood pressures of adults in the USA. Since it is normally distributed, we would apply the formula for determining z score which is expressed as
z = (mean - population mean)/standard deviation
From the information given,
population mean = 121
Standard deviation = 16
For stage 2 high blood pressure, the probability is
P(x greater than or equal to 160). It is also equal to 1 - P(x < 160)
Thus, for x = 160, we have
z = (160 - 121)/16 = 2.4375
From the standard normal distribution table, the probability value corresponding to a z score of 2.4375 is 0.9927
P(x < 160) = 0.9927
P(x greater than or equal to 160) = 1 - 0.9927 = 0.0073
Converting to percentage, it is 0.0073 * 100 = 0.73%
b) If 2000 peaople were sampled, the number of people with stage 2 high blood pressure would be
0.73/100 * 2000 14.6
To the nearest person, it is 15 people
c) For stage 1, the probability is
P(140 < x < 160)
For x = 140,
z = (140 - 121)/16 = 1.1875
From the standard normal distribution table, the probability value corresponding to a z score of 1.1875 is 0.883
Recall, for x = 160, the probaility is 0.9927
Thus,
P(140 < x < 160) = 0.9927 - 0.883 = 0.1097
Converting to percentage, it is
0.1097 * 100 = 10.97%
d) The 30th percentile refers to all values of blood pressure below k, where k is the 30th percentile. This means that we would find
P(x < k) = 0.3
The z score corresponding to a probability value of 0.3 is - 0.52
Thus,
(k - 121)/16 = - 0.52
k - 121 = - 0.52 * 16 = - 8.32
k = - 8.32 + 121
k = 112.68
The pressure for the 30th percentile is 112.68