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KatRina [158]
3 years ago
15

A normal distribution has a mean of 75 and a standard deviation of 10 use the standard normal table to find the probability that

a randomly selected X value from the distribution is less than or equal to 50
A. -2.5
B. 0.0062
C. 0.9938
D. 2.5
Mathematics
1 answer:
-BARSIC- [3]3 years ago
7 0

Answer:

0.006

Step-by-step explanation:

We are actually looking for the area under the standard normal curve to the left of 50.  The proper distribution function on my old TI calculator is normalcdf(:

normalcdf(-1000,50,75, 10),

which returns 0.006   (Answer B)

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Answer:

a) 0.3064 = 30.64% probability that athlete will test positive for prohibited drug use.

b) 0.2693 = 26.93% probability that this drug user never used the prohibited drug.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

A. If an athlete is selected at random, what is the probability that athlete will test positive for prohibited drug use?

100 - 5 = 95% of 8%

100 - 13 = 87% of 17%

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So

p = 0.95*0.08 + 0.87*0.17 + 0.11*0.75 = 0.3064

0.3064 = 30.64% probability that athlete will test positive for prohibited drug use.

B. If an athlete tests positive after a drug test, what is the probability that this drug user never used the prohibited drug?

Conditional Probability

Event A: Tests Positive

Event B: Never used the drug.

0.3064 = 30.64% probability that athlete will test positive for prohibited drug use

This means that P(A) = 0.3064

Probability of testing positive while never using the drug.

11% of 75%. So

P(A \cap B) = 0.11*0.75 = 0.0825

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0825}{0.3064} = 0.2693

0.2693 = 26.93% probability that this drug user never used the prohibited drug.

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