Question

Use your calculator to answer the questions below. Be sure to write the command you used...

Use your calculator to answer the questions below. Be sure to write the command you used and the values you entered.
Based on long-term investigations, researchers have suggested that the pH level of rainfall in the Shenandoah Mountains can be described by the Normal model ? 4.9, 0.6 . (3 points each)
a) What percent of storms produce rainfall with a pH level over 6?
b) What percent of storms produce rainfall with a pH level under 4?
c) What percent of storms produce rainfall with a pH level between 3.8 and 5.5?
  
d) The lower the pH level, the more acidic the rain. What is the pH level for the most acidic 20% of all storms?
e) What is the pH level for the least acidic 5% of all storms?
f) In what interval of pH values do the middle 60% of all storms fall?


4) A Normal model in which every value is a z-score is called The Normal Model where the mean is zero and the standard deviation is 1. Thus, to denote The Normal Model we write N(0, 1). Use The Normal Model to answer the questions below. Notice that your calculator is more exact than the 68-95-99.7 Rule. (3 points each)
a) According to the 68-95-99.7 Rule, what percent of values should have a z-score greater then 2?
b) Using your calculator, what percent of values should have a z-score greater then 2? Be sure to write the command you used and the values you entered.
c) According to the 68-95-99.7 Rule, what z-scores separate the central 95% of values?
d) Using your calculator, what z-scores separate the central 95% of values? Round your answers to two decimal places. Be sure to write the command you used and the values you entered.

Homework Answers

Answer #1

Ans

a)

z=(6-4.9)/0.6

z=1.833

P(z>1.833)=0.0334 or 3.34%

b)

z=(4-4.9)/0.6

z=-1.5

P(z<-1.5)=0.0668 or 6.68%

c)

z(3.8)=(3.8-4.9)/0.6=-1.833

z(5.5)=(5.5-4.9)/0.6=1

P(-1.833<z<1)=P(z<1)-P(z<-1.833)=0.8413-0.0334=0.8079 or 80.79%

d)

P(Z<=z)=0.2

z=normsinv(0.2)=-0.84

x=4.9-0.84*0.6

x=4.40

e)

P(Z>z)=0.05

P(Z<=z)=1-0.05=0.95

z=normsinv(0.95)=1.645

x=4.9+1.645*0.6

x=5.89

f)

z cut off values for middle 60% are +/-0.84

lower value=4.9-0.84*0.6=4.40

upper value=4.9+0.84*0.6=5.40

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