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Serjik [45]
4 years ago
9

Scores on the Common final exam in Stat 1222 course are randomly distributed with mean 75 and Standard deviation 5. The departme

nt decides to give A to all students whose scores are in the top 10% on this exam.
What is the minimum score for a student to receive A?
Mathematics
1 answer:
SSSSS [86.1K]4 years ago
4 0

Answer:

The minimum score for a student to receive A is 81.4.

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 75, \sigma = 5

The department decides to give A to all students whose scores are in the top 10% on this exam.

What is the minimum score for a student to receive A?

The minimum score is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28.

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

1.28 = \frac{X - 75}{5}

X - 75 = 1.28*5

X = 81.4

The minimum score for a student to receive A is 81.4.

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The y-intercept is y=1.

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3.2 + 6.1 minus 0.2 = ?
NemiM [27]

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Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Question on Statistics and Confidence Intervals
Digiron [165]

The phrasing "nine times out of ten" means 9/10 = 0.90 = 90% is the confidence level. We're confident 90% of the time that the confidence interval captures the population parameter we're after (in this case mu = population mean)

The portion "have an average score within 5% of 75%" means that 75% = 0.75 is the center of the confidence interval, and it goes as low as 0.75 - 0.05 = 0.70 and as high as 0.75 + 0.05 = 0.80

This confidence interval is from 70% to 80%, meaning that nine times out of ten, we're confident that the average score is between 70% and 80%

We write the confidence interval as (0.70, 0.80). It's common to use the notation (L, U) to indicate the lower (L) and upper (U) boundaries. You might see the notation in the form L < mu < U. If so, then it would be 0.70 < mu < 0.80; either way they mean the same thing.

The margin of error is 0.05 as its the 5% radius of the interval. It tells us how far the most distant score is from the center (75%)

=========================================

In summary, we have these answers

  • confidence level = 90%
  • margin of error = 5% = 0.05
  • confidence interval = (0.70, 0.80)
  • interpretation = We're 90% confident that the average exam score is between 0.70 and 0.80
4 0
3 years ago
Find the distance betwen the folowing points.
sveticcg [70]

Answer:

a. 5

b. 5

c. 6√2

d. √34

e. 43 (appx.)

Step-by-step explanation:

The distance between two points on the coordinate plane having coordinate (x1, y1) and (x2, y2) is given by \sqrt{(x1-x2)^{2}+(y1-y2)^{2}} .... (1)

a. Hence, the distance between points A(2,3) and point B(6,6) is \sqrt{(2-6)^{2}+(3-6)^{2}}=5 units. {Using equation (1)}

b. Coordinates of A and B in the Cartesian coordinate plane are (2,3) and (6,6) respectively.

Hence, the distance between A and B is \sqrt{(2-6)^{2}+(3-6)^{2}}=5 units.

c. Coordinates of A and B in the Cartesian coordinate plane are (-1,5) and (5,11) respectively.

Hence, the distance between A and B is \sqrt{(-1-5)^{2}+(5-11)^{2}}=6\sqrt{2} units.

d. Coordinates of A and B in the Cartesian coordinate plane are (1,-2) and (-2,3) respectively.

Hence, the distance between A and B is \sqrt{(1+2)^{2}+(-2-3)^{2}}=(34)^{\frac{1}{2}} units.

e. Coordinates of A and B in the Cartesian coordinate plane are (12,-12) and (-23,13) respectively.

Hence, the distance between A and B is \sqrt{(12+23)^{2}+(-12-13)^{2}}=43 units. (Approximate) (Answer)

3 0
3 years ago
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