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The scores on a statistics test are Normally distributed with parameters mean = 80 and standard deviation = 196. Find the probability that a randomly chosen score is no greater than 70

My attempt,

Pr(X < 70) = Pr(z < -0.05) = 1 - Pr(z < 0.05) = 1 - 0.5199 = .4801.

The answer given in the text book is 0.2389 tho. Could someone please help me with this.

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  • $\begingroup$ Can you explain how you got $Pr(z < -0.05)$? Not that your answer says "48% of people score below 70$, which doesn't seem to jive with "50% scored below 80". $\endgroup$
    – Calvin Lin
    Jan 19 '14 at 17:26
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    $\begingroup$ And can you check if the problem stated standard deviation, or variance? an SD of 196 seems extremely high (esp if we assume that the test scores range from 0 to 100). $\endgroup$
    – Calvin Lin
    Jan 19 '14 at 17:28
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    $\begingroup$ Yeah, if we use variance = 196 the book's answer seems to be correct. $\endgroup$
    – Tyler
    Jan 19 '14 at 17:29
  • $\begingroup$ Pr(z<−0.05), I did (70-80)/196. The books says sigma is 196. Must have been a miss print $\endgroup$ Jan 19 '14 at 17:38
  • $\begingroup$ yeah, using 196 as the variance gets all the other answers correct as well. Thanks guys $\endgroup$ Jan 19 '14 at 17:42
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Assuming that the variance is 196, then the SD is 14. The z corresponding to 70 is obtained using the standard formula (70 - mean)/SD, which in this case yields -10/14 = -0.71. Consulting the z-table, the value of 0.71 corresponds to a probability of 0.7611. Thus, the value of -0.71 corresponds to a probability of 1 - 0.7611 = 0.2389.

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