Can anyone help me find the area pls ty
2 answers:
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Answer:
0.025 = 2.5% probability that a given class period runs between 51.5 and 51.75 minutes.
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability that we find a value X lower than x is between c and d is given by the following formula
For this problem, we have that:
. So
0.025 = 2.5% probability that a given class period runs between 51.5 and 51.75 minutes.
Answer:
Step-by-step explanation:
Hello!
You have the variable
X: Area that can be painted with a can of spray paint (feet²)
The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²
since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.
a.
P(X>27)
First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)
P(Z>(27-25)/3)
P(Z>0.67)
Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of , you have to do the following conervertion:
P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143
b.
There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².
The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:
X[bar]~N(μ;δ²/n)
So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)
To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.
c.
P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999
d.
No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.
I hope it helps!
Hi there!
To solve both of you questions, you will need to use the cross product method :
<u>Finding "x" :</u>
So you have to do :
(4 × ) ÷ 2 = x gallon
4 times is the same thing as 4 times 0.5, which equals 2.
2 ÷ 2 = x gallon
1 = x gallon
So the amout for x would be : 1 gallon
<u>Finding "y" :</u>
You could do the exact same method as we did for finding "x", but the easiest and fastest way to get our answer is just by multiplying the amout we got for "x" by 2 because we are putting 8 scoops instead of 4 for "x" (and 4 times 2 = 8) :
1 gallon × 2 = y
2 gallons = y
So your answer is : The amount for y would be : 2 gallons
There you go! I really hope this helped, if there's anything just let me know! :)