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9 months ago

I need to send photos.

Mathematics

1 answer:

frez [133]9 months ago

6 0

SOLUTION:

Step 1:

From the given question, and comparing the scores in classes A and B

Step 2:

Question one, to know the class which had the better overall result on the exam;

(I) The scores in class A ranges from 65 to 100, while that of class B ranges from 60 to 90. This explains that class A had better results.

(ii) The median of the scores in class A is 85, while the median of the scores in class B is 75, this is another piece of supporting evidence that class A had better results.

Step 3:

Question two, to know the class which had greater variability in the results;

For a better understanding of this question, I need to explain the concept of variability in statistics.

Variability in statistics refers to the difference being exhibited by data points within a data set, as related to each other or as related to the mean. This can be expressed through the range, variance or standard deviation of a data set.

Step 4:

So we need to find the range of scores in each of the two classes and then compare, the class with the greater range has the greater variability.

The Range is the difference between the lowest and highest score (H - L), where H is the highest score and L is the lowest score.

Step 5:

Applying the formula for range stated in step 4;

For Class A; H = 100 and L = 65

For Class B; H = 90 and L = 60

The range for class A; H - L = 100 - 65 = 35

The range for class B; H - L = 90 - 60 = 30

By comparing the range of class A and that of B, it is clear that Class A had a greater range (variability)

CONCLUSION:

Class A had better overall results in the exam and greater variability in the results.

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3 years ago

Read 2 more answers

For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure.

Finger [1]

Answer:

Lower limit: 113.28

Upper limit: 126.72

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:

$\mu = 120, \sigma = 8$

Middle 60%

So it goes from X when Z has a pvalue of 0.5 - 0.6/2 = 0.2 to X when Z has a pvalue of 0.5 + 0.6/2 = 0.8

Lower limit

X when Z has a pvalue of 0.20. So X when $Z = -0.84$

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

$-0.84 = \frac{X - 120}{8}$

$X - 120 = -0.84*8$

$X = 113.28$

Upper limit

X when Z has a pvalue of 0.80. So X when $Z = 0.84$

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

$0.84 = \frac{X - 120}{8}$

$X - 120 = 0.84*8$

$X = 126.72$

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3 years ago

Your sister is getting married this summer and she has asked you to help her select a D.J.

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now, Marvin's Mobile Music charges 135 at first, but then is more expensive per hour is 15 bucks per each hour, so if she has a party for "h" hours, then his total charges are 135 + 15h.

now, let's check when they're both equal, at how many hours will they cost your sister the same amount.

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\\\\\\
15=4.5h\implies 15=4\frac{1}{2}h\implies 15=\cfrac{9}{2}h\implies \cfrac{15}{\frac{9}{2}}=h\implies \cfrac{\frac{15}{1}}{\frac{9}{2}}=h
\\\\\\
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so, 3 and a third of a hour, or 3 hours and 20 minutes, they both will cost your sister the same amount, so it wouldn't matter who's hired if the party is only 3 hours and 20 minutes long.

however, after that, the one with higher hourly rate, Marvin's, gets more expensive, and she's better off hiring the one with cheaper hourly rate.

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3 years ago

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viva [34]

Answer:

x=4

Step-by-step explanation:

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