Question #200552

1. A normal distribution has u=80 and o=10. What is the probability of randomly selecting the following scores?

a) x > 75

b) x < 85

c) between the mean and a score of 90

d) between the mean and a score of 50

e) 75 < x < 85

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2. Determine the z-score value in each of the following scenarios:

a) what z-score value separates the top 8% of a normal distribution from the bottom 92%?

b) what z-score value separates the top 72% of a normal distribution from the bottom 28%?

c) what z-scores values represents the middle 90% of the values in a normal distribution?


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1
Expert's answer
2021-06-02T14:12:26-0400

1.

μ=80σ=10\mu=80 \\ \sigma=10

a)

P(x>75)=1P(x<75)=1P(Z<758010)=1P(Z<0.5)=10.3085=0.8085P(x>75) = 1 -P(x<75) \\ =1 -P(Z<\frac{75-80}{10}) \\ = 1 -P(Z< -0.5) \\ = 1 -0.3085 \\ = 0.8085

b)

P(x<85)=P(Z<858010)=P(Z<0.5)=0.6914P(x<85) = P(Z< \frac{85-80}{10}) \\ = P(Z<0.5) \\ = 0.6914

c)

P(80<x<90)=P(x<90)P(x<80)=P(Z<908010)P(Z<808010)=P(Z<1)P(Z<0)=0.84130.5=0.3413P(80<x<90) =P(x<90) -P(x<80) \\ =P(Z< \frac{90-80}{10}) -P(Z< \frac{80-80}{10}) \\ = P(Z< 1) -P(Z<0) \\ = 0.8413 -0.5 \\ = 0.3413

d)

P(50<x<80)=P(x<80)P(x<50)=P(Z<808010)P(Z<508010)=P(Z<0)P(Z<3)=0.50.00135=0.49865P(50<x<80) = P(x<80) -P(x<50) \\ =P(Z< \frac{80-80}{10}) -P(Z< \frac{50-80}{10}) \\ = P(Z<0) -P(Z< -3) \\ = 0.5 -0.00135 \\ = 0.49865

e)

P(75<X<85)=P(X<85)P(X<75)=P(Z<858010)P(Z<758010)=P(Z<0.5)P(Z<0.5)=0.69140.3085=0.3829P(75<X<85) = P(X<85) -P(X<75) \\ = P(Z< \frac{85-80}{10}) -P(Z< \frac{75-80}{10}) \\ =P(Z<0.5) -P(Z< -0.5) \\ = 0.6914 -0.3085=0.3829

2.

a) P(Z<z*) = 0.92

z*=1.4051

b) P(Z<z*) = 0.28

z*=-0.5828

c)

P(z1<Z<z2)=0.900.50.902=0.05P(Z<z1)=0.05z1=1.65z2=1.65P(z^*_1<Z<z^*_2)=0.90 \\ 0.5 - \frac{0.90}{2}=0.05 \\ P(Z<z^*_1) = 0.05 \\ z^*_1=1.65 \\ z^*_2=-1.65


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