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sergij07 [2.7K]
2 years ago
8

Somebody help me with this?

Mathematics
1 answer:
trasher [3.6K]2 years ago
4 0

Answer:

Zero's

Step-by-step explanation:

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The regression equation below relates the scores students in an advanced statistics course received for homework completed and f
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Step-by-step explanation:

The homework score is an independent variable (known) and on the x-axis, while the midterm exam is the dependent variable, so it's on the y-axis

y = 0.91x - 94.4; and x=420

Substitute x as 420 into the equation.

y = 382.2 - 94.4 = 287.8, or approximately 288.

And this value isn't in the options above.

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3 years ago
Read 2 more answers
Consider the probability that exactly 90 out of 148 students will pass their college placement exams. Assume the probability tha
Pepsi [2]

Answer:

0.0491 = 4.91% probability that exactly 90 out of 148 students will pass their college placement exams.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed distributions can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

Assume the probability that a given student will pass their college placement exam is 64%.

This means that p = 0.64

Sample of 148 students:

This means that n = 148

Mean and standard deviation:

\mu = E(X) = np = 148(0.64) = 94.72

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{148*0.64*0.36} = 5.84

Consider the probability that exactly 90 out of 148 students will pass their college placement exams.

Due to continuity correction, 90 corresponds to values between 90 - 0.5 = 89.5 and 90 + 0.5 = 90.5, which means that this probability is the p-value of Z when X = 90.5 subtracted by the p-value of Z when X = 89.5.

X = 90.5

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

Z = \frac{90.5 - 94.72}{5.84}

Z = -0.72

Z = -0.72 has a p-value of 0.2358.

X = 89.5

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

Z = \frac{89.5 - 94.72}{5.84}

Z = -0.89

Z = -0.89 has a p-value of 0.1867.

0.2358 - 0.1867 = 0.0491.

0.0491 = 4.91% probability that exactly 90 out of 148 students will pass their college placement exams.

5 0
2 years ago
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