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marta [7]
2 years ago
9

Which is the correct excel formula for the 80th percentile of a distribution that is n(475, 33)? =norm.dist(80,475,33,1) =norm.i

nv(0.80,475,33) =norm.s.inv((80-475)/33)?
Mathematics
1 answer:
Lesechka [4]2 years ago
4 0

To find the 80th percentile, you have to use one of the inverse CDF functions. NORM.DIST is not one of them; it returns a probability. The 80th percentile the value of the random variable for which 80% of the distributions falls below that value.

NORM.INV(0.80, 475, 33) returns what you want.

NORM.S.INV could also work, but not with the given arguments. NORM.S.INV(0.80) alone will give you the 80th percentile for the standard normal distribution (mean 0, standard deviation 1), approximately 0.8416. From here, you would convert this value to the given distribution using

Z=\dfrac{X-\mu}\sigma\iff X=\mu+\sigma Z

so that

X=475+33(0.8416)\approx502.77

NORM.INV gives this value directly.

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The  probability   P(\^ p  >  0.80) =  0.99957

Step-by-step explanation:

From the question we are told that

     The population proportion is  p  =  0.90

     The sample size is  n  =  30

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Generally the standard deviation is mathematically represented as

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Generally the he probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places is mathematically represented as

       P(\^ p  >  0.80) =  P(\frac{ \^ p - p }{ \sigma }  >  \frac{0.80 - 0.90}{0.03} )

Generally \frac{\^p - p}{\sigma}  =  Z(The standardized \  value  \  of  \^  p)

So  

    P(\^ p  >  0.80) =  P(Z  >  -3.33 )

From the z-table P(Z  >  -3.33 ) = 0.99957

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