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Alex73 [517]
3 years ago
13

Basic Computation: Find Probabilities In Problems 5-14, assume that x has a normal distribution with the specified mean and stan

dard deviation. Find the indicated probabilities. P(3 lessthanorequalto x lessthanorequalto 6)
: mu = 4: sigma = 2 P(50 lessthanorequalto x lessthanorequalto 70)
: mu = 40: sigma = 15 P(8 lessthanorequalto x lessthanorequalto 12)
: mu = 15: sigma = 3.2 P(x greaterthanorequalto30)
: mu = 20: sigma = 3.4 P(x greaterthanorequalto90)
: mu = 100: sigma = 15 P(10 lessthanorequalto x lessthanorequalto 20)
: mu = 15: sigma = 4 P(7 lessthanorequalto x lessthanorequalto 9)
: mu = 5: sigma = 1.2 P(40 lessthanorequalto x lessthanorequalto 47)
: mu = 50: sigma = 15 p(x greaterthanorequalto 120)
: mu = 10: sigma = 15 P(x greaterthanorequalto 2)
: mu = 3
: sigma = 0.25
Mathematics
1 answer:
Ulleksa [173]3 years ago
3 0

Answer:

the answer is below

Step-by-step explanation:

The z score is used to calculate by how many standard deviations the raw score is above or below the mean. The z score is given as:

z=\frac{x-\mu}{\sigma}\\\\\mu=mean,\sigma=standard\ deviation

1) For x = 3

z=\frac{x-\mu}{\sigma}=\frac{3-4}{2}=-0.5

For x = 6

z=\frac{x-\mu}{\sigma}=\frac{6-4}{2}=1

P(3 ≤ x ≤ 6) = P(-0.5 ≤ z ≤ 1) = P(z < 1) - P(z < -0.5) = 0.8413 - 0.3085 = 0.5328

2) For x = 50

z=\frac{x-\mu}{\sigma}=\frac{50-40}{15}=0.67

For x = 70

z=\frac{x-\mu}{\sigma}=\frac{70-40}{15}=2

P(50 ≤ x ≤ 70) = P(0.67 ≤ z ≤ 2) = P(z < 2) - P(z < 0.67) = 0.9772 - 0.7486 = 0.2286

3) For x = 8

z=\frac{x-\mu}{\sigma}=\frac{8-15}{3.2}=-2.19

For x = 12

z=\frac{x-\mu}{\sigma}=\frac{12-15}{3.2}=-0.94

P(8 ≤ x ≤ 12) = P(-2.19 ≤ z ≤ -0.94) = P(z < -0.94) - P(z < -2.19) = 0.1736 - 0.0143 = 0.1593

4) For x = 30

z=\frac{x-\mu}{\sigma}=\frac{30-20}{3.4}=2.94

P(x ≥ 30) = P(z ≥ 2.94) = 1 - P(z < 2.94) = 1 - 0.9984 = 0.0016

5)  x = 90

z=\frac{x-\mu}{\sigma}=\frac{90-100}{15}=-0.67

P(x ≥ 90) = P(z ≥ -0.67) = 1 - P(z < -0.67) = 1 - 0.2514 = 0.7486

6)  For x = 10

z=\frac{x-\mu}{\sigma}=\frac{10-15}{4}=-1.25

For x = 20

z=\frac{x-\mu}{\sigma}=\frac{20-15}{4}=1.25

P(10 ≤ x ≤ 20) = P(-1.25 ≤ z ≤ 1.25) = P(z < 1.25) - P(z < -1.25) = 0.8944 - 0.1056 = 0.7888

7)  For x = 7

z=\frac{x-\mu}{\sigma}=\frac{7-5}{1.2}=1.67

For x = 9

z=\frac{x-\mu}{\sigma}=\frac{9-5}{1.2}=3.33

P(7 ≤ x ≤ 9) = P(1.67 ≤ z ≤ 3.33) = P(z < 3.33) - P(z < 1.67) = 0.9996 - 0.9525 = 0.0471

8)  For x = 40

z=\frac{x-\mu}{\sigma}=\frac{40-50}{15}=-0.67

For x = 47

z=\frac{x-\mu}{\sigma}=\frac{47-50}{15}=-0.2

P(40 ≤ x ≤ 47) = P(-0.67 ≤ z ≤ -0.2) = P(z < -0.2) - P(z < -0.67) = 0.4207 - 0.2514 = 0.1693

9)  x = 120

z=\frac{x-\mu}{\sigma}=\frac{120-10}{15}=7.33

P(x ≥ 120) = P(z ≥ 7.33) = 1 - P(z < 7.33) = 1 - 0.9999 = 0.001

10) x = 2

z=\frac{x-\mu}{\sigma}=\frac{2-3}{0.25}=-4

P(x ≥ 2) = P(z ≥ -4) = 1 - P(z < -4) = 1 - 0.0001 = 0.999

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Answer:

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Step-by-step explanation:

The diagram interpreting the question has been attached to this response.

As shown in the diagram,

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Substitute these values into equation (i) and solve as follows;

c² = (487)² + (312)² - 2(312)(487)cos(42)

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√c² = √108693.8272

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Therefore, the two airplanes are far apart by 330miles

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