Using the normal distribution, it is found that 63.18% of the area under the curve of the standard normal distribution is between z = − 0.9 z = - 0.9.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean 
and standard deviation 
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The area within 0.9 standard deviations of the mean is the <u>p-value of Z = 0.9(0.8159) subtracted by the p-value of Z = -0.9(0.1841)</u>, hence:
0.8159 - 0.1841 = 0.6318 = 63.18%.
More can be learned about the normal distribution at brainly.com/question/4079902
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<span>Moores Family Paid = 6 dollars mores for skate rental
than the cotters did.
30 dollars = the total amount they paid for skate rentals.
Let Family Moore have the variable X
Let the Family Cotters have the variable Y
=> x = y + 6
=> y = 30 – 6 / 2
First, let’s solve the Y
=> Y – 30 – 6 / 2
=> Y = 24 / 2
=> y = 12
Now, let’s solve for X
=> X = y + 6
=> x = 12 + 6
=> X = 18
</span>
Answer:
8.944271909999
Step-by-step explanation:
hope you happy
Answer:
x = -5
y = -5
Step-by-step explanation:
If you multiply the top equation by negative 2, you can add it to the bottom equation and cancel out the x:
-2(-x + 5y = -20) --> 2x - 10y = 40
Add this to the bottom equation top to bottom:
2x - 10y = 40
-2x + 4y = -10
You get:
-6y = 30
Then solve for y:
y = -5
Plug this back in for either equation and solve for x:
-x + 5(-5) = -20
-x -25 = -20
-x = 5
x = -5
Your answers are:
x = -5
y = -5
Answer:
Therefore the probability that a randomly selected student has time for mile run is less than 6 minute is 0.0618
Step-by-step explanation:
Normal with mean 6.88 minutes and
a standard deviation of 0.57 minutes.
Choose a student at random from this group and call his time for the mile Y. Find P(Y<6)
y ≈ normal (μ, σ)
The z score is the value decreased by the mean divided by the standard deviation
Therefore the probability that a randomly selected student has time for mile run is less than 6 minute is 0.0618
