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Nana76 [90]
3 years ago
13

The population mean annual salary for environmental compliance specialists is about ​$63 comma 500. A random sample of 31 specia

lists is drawn from this population. What is the probability that the mean salary of the sample is less than ​$60 comma 500​? Assume sigmaequals​$6 comma 200.4
Mathematics
1 answer:
Nat2105 [25]3 years ago
3 0

Answer:

0.0035289

Step-by-step explanation:

From the question;

mean annual salary = $63,500

n = sample size = 31

Standard deviation = $6,200

Firstly, we calculate the z-score of $60,500

Mathematically;

z-score = x-mean/SD/√n = (60500-63500)/6200/√(31) = -2.6941

So we want to find the probability that P(z < -2.6941)

We can get this from the standard normal table

P( z < -2.6941) = 0.0035289

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Step-by-step explanation:

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vladimir2022 [97]

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C) The probability the golfer got six holes-in-one during a single game is close to 0%.

<h2 /><h2><u>How to determine probabilities</u></h2>

Since a miniature golf player sinks a hole-in-one about 12% of the time on any given hole and is going to play 8 games at 18 holes each, to determine A) what is the probability the golfer got zero or one hole -in-one during a single game, B) what is the probability the golfer got exactly two holes-in-one during a single game, and C) what is the probability the golfer got six holes-in-one during a single game , the following calculations must be performed:

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Therefore, the probability the golfer got six holes-in-one during a single game is close to 0%.

Learn more about probabilities in brainly.com/question/25273534

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Step-by-step explanation:

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