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mylen [45]
3 years ago
6

4.36 Stats final scores. Each year about 1500 students take the introductory statistics course at a large university. This year

scores on the final exam are distributed with a median of 74 points, a mean of 70 points, and a standard deviation of 10 points. There are no students who scored above 100 (the maximum score attainable on the final) but a few students scored below 20 points. (a) Is the distribution of scores on this final exam symmetric, right skewed, or left skewed
Mathematics
2 answers:
Nesterboy [21]3 years ago
5 0

Answer:

Step-by-step explanation:

median 74

mean 70

this is the answer

velikii [3]3 years ago
3 0

Answer:

The distribution of scores on this final exam is left-skewed.

Step-by-step explanation:

We use the Pearson Mode Skewness to solve this question. It states that:

If the median is higher than the mean, the distribution is left-skewed.

If the median is lower than the mean, the distribution is right-skewed.

If the median is the same as the mean, the distribution is symmetric.

In this problem, we have that:

Median = 74

Mean = 70

Median higher than the mean

So the distribution of scores on this final exam is left-skewed.

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Solve the inequality please- 21 ≥ 3(a- 7) +9 Thank you so much
Alexeev081 [22]

Answer:

<h2>         a ∈ (-∞, -3></h2>

Step-by-step explanation:

<h3>- 21 ≥ 3(a - 7) + 9</h3><h3>- 21 ≥ 3a - 21 + 9</h3>

  +21         +21

<h3>    0 ≥ 3a + 9 </h3><h3>3a + 9 ≤ 0</h3>

      -9      -9

<h3>3a ≤ - 9</h3>

 ÷3    ÷3

<h3>  a ≤ -3 </h3><h3>a ∈ (-∞, -3></h3>
8 0
3 years ago
Given the following matrices A and B, find an invertible matrix U such that UA = B:
Naddik [55]

<span>A and B must be invertible, we have UA=B, since A is invertible.  A^-1 exists, by multiplying with A^-1, we have UA A^-1 =B A^-1. But AA^-1 = I (identity matrix) and XI=X, for all matrix X, we find UI= B A^-1, and U= B A^-1.</span>

3 0
2 years ago
What time does that say​
zubka84 [21]

Answer:

5:15

Step-by-step explanation:

The shorter "leg" show the hour and the longer "leg" shows the minutes. The minutes around the clock goes by 5s.

4 0
2 years ago
A grading scale is set up for 1000 students' test scores. It assumes the
astra-53 [7]

Answer:

464 students will score between 48 and 75. Using the z-distribution, we measure how many standard deviations each score is from the mean, then find the p-value associated with each score to find the proportion, and from the proportion, we find how many out of 1000.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

It assumes the scores are normally distributed with a mean score of 75 and a standard deviation of 15.

This means that \mu = 75, \sigma = 15

How many students will score between 48 and 75?

First we find the proportion, which is the pvalue of Z when X = 75 subtracted by the pvalue of Z when X = 48. So

X = 75

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

Z = \frac{75 - 75}{15}

Z = 0

Z = 0 has a p-value of 0.5

X = 48

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

Z = \frac{48 - 75}{15}

Z = -1.8

Z = -1.8 has a p-value of 0.0359

1 - 0.0359 = 0.4641

Out of 1000:

0.4641*1000 = 464

464 students will score between 48 and 75. Using the z-distribution, we measure how many standard deviations each score is from the mean, then find the p-value associated with each score to find the proportion, and from the proportion, we find how many out of 1000.

7 0
3 years ago
Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote the set of all possible pairs that can
lesya [120]

Answer:

The set A={(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(1,(2,6),(3,6),(4,6),(5,6),(6,6)}

B={(2,2),(4,2),(6,2),(2,4),(2,6),(1,3),(1,5),(1,1),(3,1),(3,3),(3,5),(4,4),(4,6),(5,1),(5,3),(5,5),(6,4),(6,6)}

C={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,5),(3,3),(3,5),(3,4),(3,6),(4,1),(4,3),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,3),(6,5)}.

A∩B:{(2,2),(4,2),(6,2),(2,4),(2,6),(4,4),(4,6),(6,4),(6,6)}

A∪B={(1,2),(2,2),(4,2),(6,2),(2,4),(2,6),(4,4),(4,6),(6,4),(6,6),(3,2),(5,2),(1,4),(3,4),(5,4),(1,6),(3,6),(5,6),(1,3),(1,5),(1,1)(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)}

A∩C={(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)}

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
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