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Musya8 [376]
3 years ago
5

PLEASE HELP

Mathematics
1 answer:
Stels [109]3 years ago
7 0

Answer:

step three is wrong

Step-by-step explanation:

just did it

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The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand
Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 18.6, \sigma = 5.9

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{21 - 18.6}{5.4}

Z = 0.44

Z = 0.44 has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768

This probability is 1 subtracted by the pvalue of Z when X = 20.4. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{20.4 - 18.6}{0.6768}

Z = 2.66

Z = 2.66 has a pvalue of 0.9961

1 - 0.9961 = 0.0039

0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

4 0
3 years ago
Hello, I am currently very stuck with this problem and I am unsure as to how I would solve it.
borishaifa [10]

We have the equation

20y=x^2-10-15

Let's complete the square, to do it let's add and subtract 25 on the right side

\begin{gathered} 20y=x^2-10-15+25-25 \\  \\ 20y=(x-5)^2-15-25_{} \\  \\ 20y=(x-5)^2-40 \\  \\  \end{gathered}

Now we can have y in function of x

\begin{gathered} y=\frac{1}{20}(x-5)^2-2 \\  \\  \end{gathered}

Now we can already identify the vertex because it's in the vertex form:

y=a(x-h)+k

Where the vertex is

(h,k)

As we can see, h = 5 and k = -2, then the vertex is

(5,-2)

Now we can continue and find the focus, the focus is

\mleft(h,k+\frac{1}{4a}\mright)

We have a = 1/20, therefore

\begin{gathered} \mleft(5,-2+5\mright) \\  \\ (5,3) \end{gathered}

The focus is

(5,3)

And the last one, the directrix, it's

y=k-\frac{1}{4a}

Then

\begin{gathered} y=-2-5 \\  \\ y=-7 \end{gathered}

Hence the correct answer is: vertex (5, -2); focus (5, 3); directrix y = -7

5 0
1 year ago
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