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Masja [62]
2 years ago
8

Kisha helped her teacher put away 36 books in boxes. She had to put the same number of books in each box to make the boxes easy

to carry. How many books could be in each box?
Mathematics
2 answers:
VashaNatasha [74]2 years ago
6 0
6 because if she put 6 books in 6 boxes, you the books would be equally split and 6 books in 6 boxes would equal 36 books
zheka24 [161]2 years ago
6 0
There could be:
1,2,3,4,6,9,12,18, or 36 books
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The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand
Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 18.6, \sigma = 5.9

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

Z = \frac{21 - 18.6}{5.4}

Z = 0.44

Z = 0.44 has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768

This probability is 1 subtracted by the pvalue of Z when X = 20.4. So

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

By the Central Limit Theorem

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

Z = \frac{20.4 - 18.6}{0.6768}

Z = 2.66

Z = 2.66 has a pvalue of 0.9961

1 - 0.9961 = 0.0039

0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

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