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Andreyy89
2 years ago
12

Adult panda weights are normally distributed with a mean of 200 pounds and a standard distribution of 20 pounds. The largest pan

das weigh over 250 pounds. Approximately what percent of the adult pandas weight over 250 pounds?
Mathematics
2 answers:
kow [346]2 years ago
5 0
250-200/20=2.5 
<span>P(2.5)=.9938 </span>
<span>that's the probability under 250 pounds. over would be 1-.9938=.0062 </span>
<span>0.62% </span>
yulyashka [42]2 years ago
3 0
To answer this problem we have to have a <span>Standard Normal Distribution table because we need to look up z scores. Then we use this equation 
</span><span>z = (value - mean)/(standard deviation) = (250-200)/20 = 2.5. 
</span><span>
The z score is then 2.5 </span>If so, look up z=2.5 on the left edge then you then determine from the table what the area to the left of that is. 
<span>
Here are the conditions: 
</span><span>p(z>2.5)=1
p(z<2.5)=.621
1-.621= .379
100-37.9= 62.1</span><span>

Given that, we should go with </span><span>.621%.</span>
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Answer:

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Step-by-step explanation:

We are given that

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By using the formula \frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}

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t=2tan(\frac{4}{5})

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P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}

We know that 1+tan^2\theta=sec^2\theta

Using the formula

P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}

P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})

By using cos^2x=\frac{1}{sec^2x}

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