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Mathematics 19 Online
OpenStudy (anonymous):

Adult panda weights are normally distributed with a mean of 200 pounds and a standard distribution of 20 pounds. The largest pandas weigh over 250 pounds. Approximately what percent of the adult pandas weight over 250 pounds?

OpenStudy (anonymous):

where do I start...

OpenStudy (anonymous):

same story, z = (value - mean)/(standard deviation) = (250-200)/20 and look up.

OpenStudy (anonymous):

I understand that but that equals a positive 2.5... @douglaswinslowcooper

OpenStudy (anonymous):

@douglaswinslowcooper Wait would it be 62.1%? (250-200)/20 = 2.5 p(z>2.5)=1 p(z<2.5)=.621 1-.621= .379 100-37.9= 62.1

OpenStudy (anonymous):

You want fraction greater than 250, which is P(z>2.5), which is (look exact value up) less than 1%.

OpenStudy (anonymous):

That is .621% @douglaswinslowcooper

OpenStudy (mathmale):

Jordi: what you've calculated (2.5) is a z-score. Suggest labeling it as such: z=2.5. Douglas wants you to look up this z score in a Standard Normal Distribution table. Have one? If so, look up z=2.5 on the left edge and determine from the table what the area to the left of that is.

OpenStudy (anonymous):

Yes, good work. 0.62% is correct.

OpenStudy (mathmale):

Nice work, Jordi, so long as you understand why you're using z=-2.5 when the z score you calculated was z=+2.5.

OpenStudy (anonymous):

OpenStudy (anonymous):

@mathmale @douglaswinslowcooper

OpenStudy (anonymous):

The answer choices are completly different

OpenStudy (anonymous):

You are looking at the tail, P(z>=2.5), which has the same probability as the other tail, P(z<=-2.5) for the normal distribution.

OpenStudy (anonymous):

So am I right? @douglaswinslowcooper

OpenStudy (anonymous):

I think you and I are right. What are your choices?

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