I need help with Part A i’ll give you brainlest if you answer this correctly!!!
1 answer:
You might be interested in
Answer:
And we can find this probability using the normal standard distribution or excel and we got:
Step-by-step explanation:
For this case we assume the following complete question: "The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.86 ounces and a standard deviation of 0.13ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places. "
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
Where and
We are interested on this probability
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability using the normal standard distribution or excel and we got:
OK first let's check the x=1.5.
Oh my, that's called a depressed cubic, no
term. There's a formula for these very much like the quadratic formula but you're probably not quite old enough for that. Anyway, 
is a solution, but that's not what they're asking. They are asking us to compare
with
and conclude
It turns out we did need all the rest of it. Save those brain cells, there's lots more math coming.
~~~~~~~~~~~~~~
I love it when the student asks for more. Here's the formula for a depressed cubic. I won't derive it here (though I did earlier today, coincidentally, but I'm probably not allowed to link to my Quora answer "what led to the discovery of complex numbers" from here). We use the trick of putting coefficients on the coefficients to avoid fractions.
has solutions
Oh my, that's called a depressed cubic, no
with
and conclude
It turns out we did need all the rest of it. Save those brain cells, there's lots more math coming.
~~~~~~~~~~~~~~
I love it when the student asks for more. Here's the formula for a depressed cubic. I won't derive it here (though I did earlier today, coincidentally, but I'm probably not allowed to link to my Quora answer "what led to the discovery of complex numbers" from here). We use the trick of putting coefficients on the coefficients to avoid fractions.
has solutions
That's pretty simple, though sometimes we end up having to take the cube roots of complex numbers, which isn't that helpful. Let's try it out on
That's so