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Setler [38]
2 years ago
15

32 is 90% of what number?

Mathematics
1 answer:
spayn [35]2 years ago
3 0

Answer:

\large\boxed{32\ is\ 90\%\ of\ 35\dfrac{5}{9}}

Step-by-step explanation:

<h3>METHOD 1:</h3>

\begin{array}{ccc}32&-&90\%\\\\x&-&100\%\end{array}\qquad\text{cross multiply}\\\\\\90x=3200\qquad\text{divide both sides by 90}\\\\x=\dfrac{3200}{90}\\\\x=\dfrac{320}{9}\\\\x=35\dfrac{5}{9}

<h3>METHOD 2:</h3>

\begin{array}{cccc}32&-&90\%&\text{divide both sides by 9}\\\\\dfrac{32}{9}&-&10\%\end{array}\\\\\\90\%+10\%=100\%\to32+\dfrac{32}{9}=32+3\dfrac{5}{9}=35\dfrac{5}{9}

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A distribution of values is normal with a mean of 220 and a standard deviation of 13. From this distribution, you are drawing sa
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The interval containing the middle-most 48% of sample means is between 218.59 to 221.41.

Step-by-step explanation:

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

Normal probability distribution

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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 distributied random variable X, with mean \mu and standard deviation \sigma, the sample means with size n 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 = 220, \sigma = 13, n = 35, s = \frac{13}{\sqrt{35}} = 2.1974

Find the interval containing the middle-most 48% of sample means:

50 - 48/2 = 26th percentile to 50 + 48/2 = 74th percentile. So

74th percentile

value of X when Z has a pvalue of 0.74. So X when Z = 0.643.

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

By the Central Limit Theorem

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

0.643 = \frac{X - 220}{2.1974}

X - 220 = 0.643*2.1974

X = 221.41

26th percentile

Value of X when Z has a pvalue of 0.26. So X when Z = -0.643

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

-0.643 = \frac{X - 220}{2.1974}

X - 220 = -0.643*2.1974

X = 218.59

The interval containing the middle-most 48% of sample means is between 218.59 to 221.41.

5 0
3 years ago
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