 Francisca Rodden

2022-01-06

A normal distribution has a mean of 80 and a standard deviation of 14. Determine the value above which 80 percent of the values will occur. Cassandra Ramirez

$\mu =80$$\sigma =14$
Let X be the random variable,
Now, $z=\frac{X-\mu }{\sigma }$
$⇒z=\frac{X-80}{14}$
${z}_{1}=\frac{{X}_{1}-80}{14}$
80%= ${X}_{1}$
$⇒P\left(z>{z}_{1}\right)=0.80$
$⇒0.5+P\left(-{z}_{1}
$⇒P\left(-{z}_{1}
$⇒P\left(-{z}_{1}
$⇒-{z}_{1}=0.84$ (from Appendix B1.)
$⇒{z}_{1}=-0.84$
Now, ${z}_{1}=\frac{{X}_{1}-80}{14}$
$⇒-0.84=\frac{{X}_{1}-80}{14}$
$⇒{X}_{1}-80=-11.76$
$⇒{X}_{1}=-11.76+80$
$⇒{X}_{1}=68.24$
80% = 68.24. Linda Birchfield

Answer: $x=z\cdot s+u$
$x=-0.8416\cdot 14+80=68.22$ karton

80% of the values will occurs above 68.24.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean $\mu$ and standard deviation , 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.
In this problem, we have that:
$\mu =80,\sigma =14$
Determine the value above which 80 percent of the values will occur.
This is the value of X when Z has a pvalue of 1-0.8 = 0.2
So this is X when Z = -0.84
$Z=\frac{X-\mu }{\sigma }\phantom{\rule{0ex}{0ex}}-0.84=\frac{X-80}{14}\phantom{\rule{0ex}{0ex}}X-80=-0.84\cdot 14\phantom{\rule{0ex}{0ex}}X=68.24$
80% of the values will occurs above 68.24.

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