Step-by-step explanation:
Hello!
The SAT math scores have a normal distribution with mean μ= 540 and standard deviation σ= 119
Eleonor scored 680 on the math part.
X: score obtained in the SAT math test
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The ACT Assessment math test has a normal distribution with mean μ= 18.2 and standard deviation σ= 3.6
Gerald took the test and scored 27.
X: score obtained in the ACT math test
a.
To standardize the values you have to use the following formula:
Z= (X-μ)/σ ~N(0;1)
For each score, you have to subtract its population mean and divide it by the standard deviation.
Eleonor score:
Z= (680-540)/119= 1.176 ≅ 1.18
Gerald Score
Z= (27-18.2)/3.6= 2.44
b.
Since both variables are very different you cannot compare the "raw" scores to know which one is higher but once both of them were standardized, you can make a valid comparison.
In this case, Eleonor's score is 1.18σ away from the mean, while Gerald's score is 2.44σ away, i.e. Gerald's score is further away from the mean score than Eleonor's so his score is higher.
c.
In this item, you are asked to find the value that divides the top 10% of the population from the bottom 90%.
Symbolically you can express it as:
P(Z>c)=0.1
or
P(Z≤c)= 0.9
The tables of standard normal distribution show accumulative probabilities of P(Z<Z₁₋α), sois best to use the second expression.
In the body of the distribution table, you have to look for a probability of 0.90 and then reach the corresponding Z value looking at the table margins. The first column shows the integer and first decimal digit, the first row shows the second decimal digit. So the corresponding value of Z is 1.28
Now you have to reverse the standardization to know the corresponding scores for each test.
SAT test score:
Z= (c-μ)/σ
Z*σ = c-μ
c = (Z*σ ) + μ
c= (1.28*119)+540
c= 692.32
The student should score 692.32 in his SAT math test to be in the top 10% of the population.
ACT test score
Z= (c-μ)/σ
Z*σ = c-μ
c = (Z*σ ) + μ
c= (1.28*3.6)+18.2
c= 22.808 ≅ 22.81
The student should score 22.81 in his ACT math test to be in the top 10% of the population.
d.
In this item, their vas a sample of students that took the ACT taken and you need to calculate the probability of the sample mean being greater than 25.
If you were to take a 100 random samples of ACT scores of 100 students and calculate the mean of all of them, you will get that the sample mean is a random variable with the same kind of distribution as the original variable but it's variance will be influenced by the sample size. In this case, the original variable is:
X: score obtained in the ACT math test
This variable has a normal distribution X~N(μ;δ²), then it's the sample mean will also have a normal distribution with the following parameters X[bar]~N(μ;δ²/n)
Remember when you standardize a value of the variable of interest you subtract its "mean" and divide it by its "standard deviation" in this case the mean is μ and the standard deviation will be √(δ²/n) ⇒ δ/√n and the formula of the standard normal is:
Z= (X[bar]-μ)/(δ/√n)~N(0;1)
with n=100
μ= 18.2
δ= 3.6
P(X[bar]>25)= 1 - P(X[bar]≤25)
1 - P(Z≤(25-18.2)/(6.3/√100))= 1 - P(Z≤10.79)= 1 - 1 = 0
The probability of the mean ACT score for a random sample of 100 students being more than 25 is zero.
I hope it helps!
