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dolphi86 [110]
3 years ago
7

The median is the same as the ______________.

Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
7 0

Answer: Option A

The median is the same as the _ Second quartile_.

Step-by-step explanation:

Given a series of data ordered from least to greatest, the median is the value that is in the center.

That is, the median represents the value that divides 50% of the data.

In the same way, the first quartile Q_1 is the value that divides 25% of the data and the third quartile Q_3 is the value that divides 75% of the data.

The second quartile Q_2 is the number that divides 50% of the data.

Then notice that the second quartile is equal to the median

Then te answer is the option A.

xeze [42]3 years ago
3 0

Answer:

A. Second quartile

Step-by-step explanation:

Median is the middle value of the data when arranged, it divides the data in two halves.

Also second quartile is the point in data that divides the data into two quarters.

Hence second quartile is median of data!

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What is 9| - z/3| + 6 &lt; 30
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9(¹/₃z) + 6 < 30    or    9(¹/₃z) + 6 > -30
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8 0
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a sample of 546 boys aged 6–11 was weighed, and it was determined that 89 of them were overweight. A sample of 508 girls aged 6–
AleksAgata [21]

Answer:

p-value:  0.6527

Step-by-step explanation:

Hello!

You have two samples to study, from each sample the weight of each child was measured and counted the total of overweight kids (x: "success") in each group:

Sample 1 (Boys aged 6-11)

n₁= 546

x₁= 89

^p₁= x₁/n₁ = 89/546 ≅0.16

Sample 2 (girls aged 6-11)

n=508

x= 74

^p= x/n = 74/508 ≅ 0.15

If the hypothesis statement is "The proportion of boys that are overweight differs from the proportion of girls that are overweight", the test hypothesis is:

H₀: ρ₁ = ρ₂

H₁: ρ₁ ≠ ρ₂

This type of hypothesis leads to a two-tailed rejection region, then the p-value will also be two-tailed. To calculate the p-value you have to first calculate the value of the statistic under the null hypothesis, in this case, is a test for the difference between two proportions:

Z=<u>      (^ρ₁ - ^ρ₂) - (ρ₁ - ρ₂)        </u> ≈ N(0;1)

    √(ρ` * (1 - ρ`) * (1/n₁ + 1/n₂))

ρ`=<u> x₁ + x₂  </u> = <u>  89+74     </u> = 0.154 ≅ 0.15

     n₁ + n₂     546 + 508

Z⁰ᵇ =<u>          (0.16-0.15) - (0)                    </u>

       √(0.15 * (1 - 0.15) * (1/546 + 1/508))

Z⁰ᵇ = 0.45

I've mentioned before that in this test you have a two-tailed p-value. The value calculated (0.45) corresponds to the right or positive tail and the left tail is symmetrical to it concerning the distribution mean, in this case, is 0, so it is -0.45. To obtain the p-value you need to calculate the probability of both values and add them:

P(Z>0.45) + P(Z<-0.45) = (1- P(Z<0.45)) + P(Z<-0.45) = (1-0.67364) + 0.32636 = 0.65272 ≅ 0.6527

p-value:  0.6527

Since there is no signification level in the problem, I'll use the most common to reach a decision. α: 0.05

Since the p-value is greater than α, you do not reject the null Hypothesis, in other words, there is no significative difference between the proportion of overweight boys and the proportion of overweight girls.

I hope it helps!

3 0
3 years ago
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