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What is the probability that a data value in a normal distribution is between a z-score of -0.28 and a z-score of 0.64? Round your answer to the nearest tenth of a percent. A. 90.9% B. 58.1% C. 34.9% D. 73.3%

User John Vint
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1 Answer

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Answer:


P(-0.28 <Z< 0.64) = P(Z<0.64) -P(Z<-0.28)


P(Z<0.64) =0.739


P(Z<-0.28) =0.390

And with the difference we got: 0.739-0.390 = 0.349

And if we convert the probability to % we got 0.349*100 = 34.9%. And the best answer would be:

C. 34.9%

Explanation:

For this case we want this probability:


P(-0.28 <Z< 0.64)

And we can find this probability using the normal standard distribution and with the following difference:


P(-0.28 <Z< 0.64) = P(Z<0.64) -P(Z<-0.28)

If we find the individual probabilities we got:


P(Z<0.64) =0.739


P(Z<-0.28) =0.390

And with the difference we got: 0.739-0.390 = 0.349

And if we convert the probability to % we got 0.349*100 = 34.9%. And the best answer would be:

C. 34.9%

User Mustafa Deniz
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