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The scores received by juniors on the math portion of the PSAT are normally distributed with a mean of 48.6 and a standard deviation of 11.4. What is the probability that a randomly selected score is at least 76?

Respuesta :

Rau7star

Answer:

0.82%

Step-by-step explanation:

Given:

mean 'μ'= 48.6

standard deviation 'σ' = 11.4

raw score 'x'= 76

Z- value can be determined by:

z = (x-μ)/σ

z= (76-48.6)/11.4

z= 2.4

Now, by looking upto the z-score chart

(z<2.4)= 0.9918

since randomly selected score is at least 76, subtract the value by 1

so,

1-0.9918 = 0.0082

Converting into percentage

0.0082 x 100%

0.82%

jmonterrozar

Answer:

0.82%

Step-by-step explanation:

We have that the mean (m) is equal to 48.6, the standard deviation (sd) 11.4

They ask us for P (x >76)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / sd

We have all these values, replacing we have:

z = (76 - 48.6) / (11.4)

z = 2.4

With the normal distribution table (attached), we have that at that value, the probability is:

P (z <2.4) = 0.9918

Thus:

P (x> 76) = 1 - P (x <76)

P (x> 76) = 1 - 0.9918 = 0.0082

It means that the probability is 0.82%

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