192.5=1/2(26+12.5)h. What is the unknown measure

Mathematics

1 answer:

hram777 [196]2 years ago

8 0

$\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\[-0.5em] \hrulefill\\ a=12.5\\ b=26\\ A=192.5 \end{cases}\implies 192.5=\cfrac{h(12.5+26)}{2} \\\\\\ 385=h(12.5+26)\implies 385=38.5h\implies \cfrac{385}{38.5}=h\implies 10=h$

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Using the normal distribution and the central limit theorem, it is found that the interval that contains 99.44% of the sample means for male students is (3.4, 3.6).

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

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In this problem:

The mean is of $\mu = 3.5$ .

The standard deviation is of $\sigma = 0.5$ .

Sample of 100, hence $n = 100, s = \frac{0.5}{\sqrt{100}} = 0.05$

The interval that contains 95.44% of the sample means for male students is between Z = -2 and Z = 2, as the subtraction of their p-values is 0.9544, hence:

Z = -2:

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