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.

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Cassandra Ramirez · Accepted Answer

μ=80,&amp;nbsp;σ=14.&amp;nbsp;Let X be the random variable,&amp;nbsp;Now,&amp;nbsp;z=X−μσ&amp;nbsp;⇒z=X−8014&amp;nbsp;&amp;nbsp;z1=X1−8014&amp;nbsp;80%=&amp;nbsp;X1.&amp;nbsp;⇒P(z&amp;gt;z1)=0.80&amp;nbsp;⇒0.5+P(−z1&amp;lt;z&amp;lt;0)=0.80&amp;nbsp;⇒P(−z1&amp;lt;z&amp;lt;0)=0.80−0.5&amp;nbsp;⇒P(−z1&amp;lt;z&amp;lt;0)=0.30&amp;nbsp;⇒−z1=0.84&amp;nbsp;(from Appendix B1.)&amp;nbsp;⇒z1=−0.84&amp;nbsp;Now,&amp;nbsp;z1=X1−8014&amp;nbsp;⇒−0.84=X1−8014&amp;nbsp;⇒X1−80=−11.76&amp;nbsp;⇒X1=−11.76+80&amp;nbsp;⇒X1=68.24&amp;nbsp;80% = 68.24.

Linda Birchfield · Accepted Answer

Answer: x=z⋅s+u   x=−0.8416⋅14+80=68.22

karton · Accepted Answer

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 μ and standard deviation , the zscore of a measure X is given by: Z=X−μσ 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: μ=80,σ=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=X−μσ−0.84=X−8014X−80=−0.84⋅14X=68.24 80% of the values will occurs above 68.24.

A normal distribution has a mean of 80 and a

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