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AlekseyPX
3 years ago
15

The question says: Paula started reading a book and red 1/12 of the book on the first day. the next day, she read 1/3 of the boo

k.
How much does she need ot read to finish the book? Hint: Not 5/12.​

Mathematics
1 answer:
tester [92]3 years ago
4 0

Answer:

Step-by-step explanation:

ℎ

\frac{5}{12}

ℎ ℎ

\\  \frac{1}{12}  +  \frac{1}{3}  =  \frac{5}{12}

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The weight of crackers in a box is stated to be 16 ounces. The amount that the packaging machine puts in the boxes is believed t
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Answer:

99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces

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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 16.15, \sigma = 0.3, n = 50, s = \frac{0.3}{\sqrt{50}} = 0.0424

What is the probability that the mean weight of a 50-box case of crackers is above 16 ounces?

This is 1 subtracted by the pvalue of Z when X = 16. So

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

By the Central Limit Theorem

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

Z = \frac{16 - 16.15}{0.0424}

Z = -3.54

Z = -3.54 has a pvalue of 0.0001

1 - 0.0001 = 0.9999

99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces

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3 years ago
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