Question

Suppose systolic blood pressure of 16-year-old females is approximately normally distributed with a mean of 118 mmHg and a variance of 421.07 mmHg. If a random sample of 16 girls were selected from the population, find the following probabilities:

a) The mean systolic blood pressure will be below 119 mmHg.
probability =

b) The mean systolic blood pressure will be above 124 mmHg.
probability =

c) The mean systolic blood pressure will be between 112 and 120 mmHg.
probability =

d) The mean systolic blood pressure will be between 107 and 113 mmHg.
probability =

Answer 1

Solution :

Given that ,

mean = = 118

standard deviation = = 20.52

= / n = 20.52 / 16 = 5.13

a)

P( < 119) = P(( - ) / < (119 - 118) / 5.13)

= P(z < 0.19)

= 0.5753

Probability = 0.5753

b)

P( > 124) = 1 - P( < 124)

= 1 - P[( - ) / < (124 - 118) / 5.13]

= 1 - P(z < 1.17)

= 1 - 0.879

= 0.121

Probability = 0.121

c)

= P[(112 - 118) /5.13 < ( - ) / < (120 - 118) / 5.13)]

= P(-1.17 < Z < 0.39)

= P(Z < 0.39) - P(Z < -1.17)

= 0.6517 - 0.121

= 0.5307

Probability = 0.5307

d)

= P[(107 - 118) / 5.13< ( - ) / < (113 - 118) / 5.13)]

= P(-2.14 < Z < -0.97)

= P(Z < -0.97) - P(Z < -2.14)

= 0.166 - 0.0162

= 0.1498

Probability = 0.1498