Respuesta :
.621 percent should be your answerTo answer this problem we have to have a Standard Normal Distribution table because we need to look up z scores. Then we use this equation
z = (value - mean)/(standard deviation) = (250-200)/20 = 2.5.
The z score is then 2.5 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.
Here are the conditions:
p(z>2.5)=1
p(z<2.5)=.621
1-.621= .379
100-37.9= 62.1
Given that, we should go with .621%.
z = (value - mean)/(standard deviation) = (250-200)/20 = 2.5.
The z score is then 2.5 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.
Here are the conditions:
p(z>2.5)=1
p(z<2.5)=.621
1-.621= .379
100-37.9= 62.1
Given that, we should go with .621%.
Answer: 0.62%
Step-by-step explanation:
Given: The means weight of the adult panda:
[tex]\mu=200[/tex]
Standard deviation [tex]\sigma= 20[/tex]
Since, [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x = 250
[tex]z=\dfrac{250-200}{20}\\\\\Rightarrow\ z =2.5[/tex]
Now, the probability of the adult pandas weight less than 250 pounds
[tex]P(z)=P(2.5)= 0.9937903[/tex]
Now, the probability of the adult pandas weight over than 250 pounds
[tex]1-P(z)=1-P(2.5)=1- 0.9937903=0.0062097=0.62097\approx0.62\%[/tex]