Suppose a clinically accepted value for mean systolic blood pressure in males ag

Question

Suppose a clinically accepted value for mean systolic blood pressure in males aged 20 - 24 is 120 mm of Hg and the standard deviation is 20 mm of Hg. If a 22 year old male is selected, at random from the population, what is the probability that his systolic blood pressure is equal's or less than 150 mm of Hg? Equal to or less-than 110? The systolic blood pressure of a 20 year old male selected at random from the population was 160. How many standard deviations above the mean is this value? Between what two blood pressure readings will 95% of all systolic blood pressure readings for 20 - 24 year old males lie? Between what two values will 90% of the readings lie? What proportion of readings lie within the range 100 - 140 mm Hg? What proportion will lie outside the range 60-180 mm of Hg? As part of a screening project, the systolic blood pressures of 100 freshmen medical students were taken. The mean systolic blood pressure of this sample was 126 mm of Hg. For n= 100, what is the probability of obtaining a sample mean value equal to or greater than 126 mm of Hg from a population with true mean 120 mm of Hg? How many standard deviations above the true mean is the observed mean value? Between what two mean values would we expect 95% of all sample means (of size n 100) to lie? Based on the above information and computations, what conclusions might be drawn concerning the systolic blood pressure of medical students? Above what value would 10% of the sample means (n - 100) lie? What is the probability of selecting a sample with mean equal to less than 115 mm of Hg from a population with true mean systolic blood pressure equal to 120 mm of Hg? Place a 99% confidence interval on the true mean systolic blood pressure for all freshmen medical students

Explanation / Answer

a.
Mean ( u ) =120
Standard Deviation ( sd )=20
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X < 150) = (150-120)/20
= 30/20= 1.5
= P ( Z <1.5) From Standard Normal Table
= 0.9332                  
P(X < 110) = (110-120)/20
= -10/20= -0.5
= P ( Z <-0.5) From Standard Normal Table
= 0.3085
c.
P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.96
P( x-u/s.d < x - 120/20 ) = 0.025
That is, ( x - 120/20 ) = -1.96
--> x = -1.96 * 20 + 120 = 80.8007                  
P ( Z > x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is 1.96
P( x-u/ (s.d) > x - 120/20) = 0.025
That is, ( x - 120/20) = 1.96
--> x = 1.96 * 20+120 = 159.1993  
95% of values between [ 80.8007, 159.1993]
P ( Z < x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is -1.645
P( x-u/s.d < x - 120/20 ) = 0.05
That is, ( x - 120/20 ) = -1.64
--> x = -1.64 * 20 + 120 = 87.1029                  
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.6449
P( x-u/ (s.d) > x - 120/20) = 0.05
That is, ( x - 120/20) = 1.6449
--> x = 1.6449 * 20+120 = 152.8971                  

95% of values between [ 87.1029   ,152.8971   ]
d.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 100) = (100-120)/20
= -20/20 = -1
= P ( Z <-1) From Standard Normal Table
= 0.15866
P(X < 140) = (140-120)/20
= 20/20 = 1
= P ( Z <1) From Standard Normal Table
= 0.84134
P(100 < X < 140) = 0.84134-0.15866 = 0.6827                  
d.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 60) = (60-120)/20
= -60/20 = -3
= P ( Z <-3) From Standard Normal Table
= 0.00135
P(X < 180) = (180-120)/20
= 60/20 = 3
= P ( Z <3) From Standard Normal Table
= 0.99865
P(60 < X < 180) = 0.99865-0.00135 = 0.9973                  

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