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navik [9.2K]
4 years ago
11

In a large statistics class, the correlation between midterm scores and final scores is found to be nearly 0.50, every term. The

scatter diagrams are football-shaped. Predict the percentile rank on the final for a student whose percentile rank on the midterm is a) 5% b) 80% c) 50% d) unknown

Mathematics
1 answer:
xenn [34]4 years ago
6 0

Answer:

A) 21%

B) 66%

C) 50%

D) 50%

Step-by-step explanation:

A) We want to find the percentile rank on the final for a student whose percentile rank on the midterm is 5%.

Now, it means we are looking for the value of z on the 5th percentile.

Now from the Z-score percentile normal distribution table i have attached, the value of z on the 5th percentile is; Zx = -1.645

Now, using the regression method, we know that;

Zy = r(Zx)

Where Zy is the score at which the final percentile occurs.

And r = 0.5 from the question

Thus; Zy = 0.5 x -1.645

Zy = - 0.8225

Now looking at the table attached, there is no direct value of -0.8225,and so we pick the closest to it which is -0.8 and it falls on the 21st percentile.

Thus, the percentile rank on the final for a student whose percentile rank on the midterm is 5% would be 21%.

B) We want to find the percentile rank on the final for a student whose percentile rank on the midterm is 80%.

Following same pattern in A above;

From table, Zx = 0.842

And Zy = 0.5 x 0.842 = 0.421

The percentile for this value is approximately 66th percentile.

Thus, the percentile rank on the final for a student whose percentile rank on the midterm is 80% would be 66%.

C)We want to find the percentile rank on the final for a student whose percentile rank on the midterm is 50%.

Following same pattern in A above;

From table, Zx = 0

And Zy = 0.5 x 0 = 0

The percentile for this value is 50th percentile.

Thus, the percentile rank on the final for a student whose percentile rank on the midterm is 50% would be 50%.

D) If the score is unknown, the best bet will be to use a middle percentile of 50%.

It thus follows the same procedure from c above and same answer of 50%.

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We need to use the formula for simple interest which is
I= prt
Where I is the amount of money you earned or pay in interest
p is the principal, the amount you deposited or borrowed
r is the interest rate expressed as a decimal
t is time in terms of years

In this problem, I= 1,680
p= 3000
t= 8
'. r is what we are looking for.

Substituting the numbers into the simple interest formula, we get
I=. p r t
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1,680= 24,000r Divide both sides by 24,000
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Can 0.1/100 be simplified?
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According to a recent​ publication, the mean price of new mobile homes is ​$63 comma 800. Assume a standard deviation of ​$7900.
wolverine [178]

Answer:

a. For n=25, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are ​$63,800 and ​$1,580​, respectively.

b. For n=50, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are ​$63,800 and ​$1,117​, respectively.

Step-by-step explanation:

In this case, for each sample size, we have a sampling distribution (a distribution for the population of sample means), with the following parameters:

\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{n}

For n=25 we have:

\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{25}=7,900/5=1,580

The spread of the sampling distribution is always smaller than the population spread of the individuals. The spread is smaller as the sample size increase.

This has the implication that is expected to have more precision in the estimation of the population mean when we use bigger samples than smaller ones.

If n=50, we have:

\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{50}=7,900/7.07=1,117

4 0
4 years ago
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