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

Beginner2022-01-07Added 30 answers

$\mu =80$, $\sigma =14$.

Let X be the random variable,

Now, $z=\frac{X-\mu}{\sigma}$

$\Rightarrow z=\frac{X-80}{14}$

$z}_{1}=\frac{{X}_{1}-80}{14$

80%= $X}_{1$.

$\Rightarrow P\left(z>{z}_{1}\right)=0.80$

$\Rightarrow 0.5+P(-{z}_{1}<z<0)=0.80$

$\Rightarrow P(-{z}_{1}<z<0)=0.80-0.5$

$\Rightarrow P(-{z}_{1}<z<0)=0.30$

$\Rightarrow -{z}_{1}=0.84$ (from Appendix B1.)

$\Rightarrow {z}_{1}=-0.84$

Now, $z}_{1}=\frac{{X}_{1}-80}{14$

$\Rightarrow -0.84=\frac{{X}_{1}-80}{14}$

$\Rightarrow {X}_{1}-80=-11.76$

$\Rightarrow {X}_{1}=-11.76+80$

$\Rightarrow {X}_{1}=68.24$

80% = 68.24.

Linda Birchfield

Beginner2022-01-08Added 39 answers

Answer: $x=z\cdot s+u$

$x=-0.8416\cdot 14+80=68.22$

karton

Expert2022-01-11Added 613 answers

Answer:

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

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:

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

80% of the values will occurs above 68.24.

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