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True [87]
3 years ago
9

In a math class, 12 people reported their grades on the last test. Here are their responses:

Mathematics
1 answer:
Murrr4er [49]3 years ago
5 0

Answer:

(a) 23, 35, 28, 33, 5, 12, 40, 25, 20, 18, 1, 16

1, 5, 12, 16, 18, 20 23, 25, 28, 33, 35, 40

n = 12 M = (20 + 23)/ 2 = 21.5

Q1 = (12 + 16) / 2 = 14 Q3 = (28 + 33) /2 = 30.5

(b) 22, 33, 25, 28, 5, 12, 35, 23, 20, 18, 1, 40, 16

1, 5, 12, 16, 18, 20 | 22 | 23, 25, 28, 33, 35, 40

n = 13 M = 22 Q1 = 14 Q3 = 30.5

(c) 20, 28, 23, 25, 3, 5, 33, 22, 18, 16, 40, 1, 35, 12

1, 3, 5, 12, 16, 18, 20, 22, 23, 25, 28, 33, 35, 40

n = 14 M = (20 + 22)/2 = 21 Q1 = 12 Q3 = 28

(d) 20, 28, 23, 25, 3, 5, 30, 22, 18, 40, 16, 35, 1, 33, 12

1, 3, 5, 12, 16, 18, 20 | 22 | 23, 25, 28, 30, 33, 35, 40

n = 15 M = 22 Q1 = 12 Q3 = 30

hope this helps :)

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An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0
zmey [24]

Answer:

96.33% probability that everyone who appears for the flight will get a seat

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

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:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples 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)}.

In this problem, we have that:

n = 105, p = 0.9

So

\mu = E(X) = np = 105*0.9 = 94.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{105*0.9*0.1} = 3.07

What is the probability that everyone who appears for the flight will get a seat

100 or less people appearing to the flight, which is the pvalue of Z when X = 100. So

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

Z = \frac{100 - 94.5}{3.07}

Z = 1.79

Z = 1.79 has a pvalue of 0.9633

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She had 250 in about 13 weeks

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Answer:

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Step-by-step explanation:

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Answer:

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