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Darya [45]
3 years ago
13

An English teacher needs to pick 13 books to put on her reading list for the next school year, and she needs to plan the order i

n which they should be read. She has narrowed down her choices to 5 novels, 7 plays, 7 poetry books, and 5 nonfiction books. Step 1 of 2 : If she wants to include no more than 2 nonfiction books, how many different reading schedules are possible? Express your answer in scientific notation rounding to the hundredths place.
Mathematics
1 answer:
olga nikolaevna [1]3 years ago
4 0

Answer:

YoUvE GoT HaKeD

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A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true
kvv77 [185]

Answer:

99.74% probability that the sample proportion will be less than 0.1

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 = 276, p = 0.06

So

\mu = E(X) = np = 276*0.06 = 16.56

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{276*0.06*0.94} = 3.9454

What is the probability that the sample proportion will be less than 0.1

This is the pvalue of Z when X = 0.1*276 = 27.6. So

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

Z = \frac{27.6 - 16.56}{3.9454}

Z = 2.8

Z = 2.8 has a pvalue of 0.9974

99.74% probability that the sample proportion will be less than 0.1

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Step-by-step explanation:

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