Question

For males in a certain town, the systolic blood pressure is normally distributed with a mean of 135 and a standard deviation of 6. What percentage of males in the town that have a systolic blood pressure higher than 137, to the nearest tenth?

Answer 1

Answer:

.4

Step-by-step explanation:

N(135,6)

• standardize: z=(x-μ)/σ
• z=(137-135)/6= 0.33
• z=0.33

N(1,0)

P(Z$\geq$0.33) = 1 - P(Z$\leq$0.33)

                 = 1 - .6293

                 = 0.3707

P(Z$\geq$0.33) = 0.3707

Rounded: 0.4