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alexgriva [62]
3 years ago
12

A teacher surveys a sample of students from Lake Middle School. He asks the students where they’d like to go for a field trip. H

e records these results:
80 choose the aquarium.
60 choose the science center.
30 choose the planetarium.
40 choose the farm.
The teacher wants to figure out the relative frequency of a student choosing the aquarium. He will then estimate how many students out of all 1000 at the school are likely to choose the aquarium.
Answer the questions to solve the problem.

1. How many students did the teacher survey?
Mathematics
2 answers:
Ira Lisetskai [31]3 years ago
5 0

Answer:

210 students

Step-by-step explanation:

80 students chose the aquarium.

60 students chose the science center.

30 students chose the planetarium.

40 students chose the farm.

The students can't vote for more than one place, so you just add all the numbers together.

80 + 60 + 30 + 40 = 210 students

defon3 years ago
5 0
80+60+30+40=210 !!!!!!!!!!!!!
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Answer: −10

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3 years ago
Consider the probability that no less than 95 out of 152 registered voters will vote in the presidential election. Assume the pr
nikdorinn [45]

Answer:

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.

Step-by-step explanation:

Binomial probability distribution

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Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

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This means that n = 152

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\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{152*0.61*0.39} = 6.01

Probability that no less than 95 out of 152 registered voters will vote in the presidential election.

This is, using continuity correction, P(X \geq 95 - 0.5) = P(X \geq 94.5), which is 1 subtracted by the pvalue of Z when X = 94.5. So

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Z = \frac{94.5 - 92.72}{6.01}

Z = 0.3

Z = 0.3 has a pvalue of 0.6179

1 - 0.6179 = 0.3821

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.

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