The distribution is normal, the mean is 500 and the standard deviation is 100. What percent...

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math Statistics-And-Probability

Answer 1

Answer #1

Ans. Let X be a random variable, has N(500, 100²)

P(600<X<700)

= P[(600-500)/100 < (X-500)/100 < (700-500)/100]

= P(1 < Z < 2), where Z is standard normal variate

= P(0 < Z <2) - P(0 < Z < 1)

= 0.4772 - 0.3413

= 0.1359 = 13.59%

Thus the percentage of counts will lie between 600 and 700

is 13.59

Answer 2

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