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andrew11 [14]
3 years ago
14

A community college has a math placement exam that has a mean of 65 and a standard deviation of 8.2. If a student takes the exam

and scores in the lower 2.5%, they are eligible for a free intensive course to help them succeed in their math class. Assume that the exam has a normal distribution. What is the score that cuts off the bottom 2.5%
Mathematics
2 answers:
Marat540 [252]3 years ago
8 0

Answer:

The score that cuts off the bottom 2.5% is 48.93.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

\mu = 65, \sigma = 8.2

What is the score that cuts off the bottom 2.5%

This is X when Z has a pvalue of 0.025, so X when Z = -1.96.

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

-1.96 = \frac{X - 65}{8.2}

X - 65 = -1.96*8.2

X = 48.93

The score that cuts off the bottom 2.5% is 48.93.

Umnica [9.8K]3 years ago
6 0

Answer:

The score that cuts off the bottom 2.5% is 48.93.

Step-by-step explanation:

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vagabundo [1.1K]

Answer:

I can not answer for some reason

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Step-by-step explanation:

6 0
2 years ago
Does anyone know how to do this ??? <br> -view attachment
cricket20 [7]

As soon as I read this, the words "law of cosines" popped
into my head.  I don't have a good intuitive feeling for the
law of cosines, but I went and looked it up (you probably
could have done that), and I found that it's exactly what
you need for this problem.

The "law of cosines" relates the lengths of the sides of any
triangle to the cosine of one of its angles ... just what we need,
since we know all the sides, and we want to find one of the angles.

To find angle-B, the law of cosines says

       b² = a² + c² - 2 a c cosine(B)

B  =  angle-B
b  =  the side opposite angle-B = 1.4
a, c = the other 2 sides = 1 and 1.9

                  (1.4)² = (1)² + (1.9)² - (2 x 1 x 1.9) cos(B)

                 1.96  =  (1) + (3.61)  -  (3.8) cos(B)

Add  3.8 cos(B)  from each side:

                 1.96 + 3.8 cos(B) = 4.61

Subtract  1.96  from each side:

                             3.8 cos(B) =  2.65

Divide each side by  3.8 :

                                  cos(B)  =  0.69737  (rounded)

Whipping out the
trusty calculator:
                                 B  = the angle whose cosine is 0.69737

                                      =  45.784° .

Now, for the first time, I'll take a deep breath, then hold it
while I look back at the question and see whether this is
anywhere near one of the choices ...

By gosh !  Choice 'B' is  45.8° !                    yay !
I'll bet that's it !

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If an equation has a negative discriminant, than there are no real solutions. We know this because the quadratic formula requires you to take the square root of the discriminant and therefore you would have all imaginary numbers.

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