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weqwewe [10]
3 years ago
6

Given a normal distribution with a mean of 128 and a standard deviation of 15, what percentage of values is within the interval

83 to 173?
A. 32%
B. 50%
C. 68%
D. 95%
E. 99.7%
Mathematics
1 answer:
tatyana61 [14]3 years ago
6 0

Answer:

E. 99.7%

Step-by-step explanation:

Calculate each z-score.

z = (x − μ) / σ

z = (173 − 128) / 15

z = 3

z = (x − μ) / σ

z = (83 − 128) / 15

z = -3

You can use a z-score table, or use the empirical rule.

P(-3 < z < 3) = 99.7%.

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Answer the following questions and round your answers to 2 decimal places. 84% of all Americans live in cities with population g
nignag [31]

Answer:

15.23% probability that exactly 42 of them live in cities with population greater than 50,000 people.

Step-by-step explanation:

For each American, there are only two possible outcomes. Either they live in cities with population greater than 50,000 people. Or they do not. The probability of a person living in a city with population greater than 50,000 people is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

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This means that p = 0.84

If 50 Americans are selected at random, Find the probability that Exactly 42 of them live in cities with population greater than 50,000 people.

This is P(X = 42) when n = 50. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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3 years ago
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