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what does On May 17, 1954, Chief Justice Earl Warren issued the Supreme Court’s unanimous decision in Brown v. Board of Educatio

n, ruling that racial segregation in public schools violated the Equa

1 answer:

S_A_V [24]2 years ago

4 0

Answer:

<h3>Plessy v. Ferguson case.</h3>

Explanation:

The Plessy v. Ferguson case of 1896 ruled out that segregation was constitutional as long as both blacks and whites enjoyed opportunities that were separate but equal.

However, the Supreme Court's verdict in the Plessy v. Ferguson case of 1896 was challenged by Justice Earl Warren stating that the idea of ‘separate but equal’ in public school was unconstitutional and inherently unequal.

Thus, Chief Justice Earl Warren issued the Supreme Court’s unanimous decision in Brown v. Board of Education, ruling that racial segregation in public schools violated the "Equal Protection Clause” of the 14th Amendment.

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company manufactured six television sets on a given day, and these TV sets were inspected for being good or defective. The resul

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Sampling distribution involves the proportions of a data element in a given sample.

<em>The proportion of Good TV set is 0.67</em>
<em>The number of ways of selecting 5 from 6 TV sets is 6</em>
<em>The number of ways of selecting 4 from 6 TV sets is 15</em>

<em />

Given

$n = 6$

Sample Space = Good, Good, Defective, Defective, Good, Good

<u>(a) Proportion that are good</u>

From the sample space, we have:

$Good = 4$

So, the proportion (p) that are good are:

$p = \frac{Good}{n}$

$p = \frac{4}{6}$

$p = 0.67$

<u>(b) Ways to select 5 samples (without replacement)</u>

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 5$

So, we have:

$^6C_5 = \frac{6!}{(6 - 5)!5!}$

$^6C_5 = \frac{6!}{1!5!}$

$^6C_5 = \frac{6 \times 5!}{1 \times 5!}$

$^6C_5 = \frac{6}{1}$

$^6C_5 = 6$

Hence, there are 6 ways

<u>(c) All possible sample space of 4</u>

First, we calculate the number of ways to select 4.

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 4$

So, we have:

$^6C_4 = \frac{6!}{(6 - 4)!4!}$

$^6C_4 = \frac{6!}{2!4!}$

$^6C_4 = \frac{6 \times 5 \times 4}{2 \times 1 \times 4!}$

$^6C_4 = \frac{30}{2}$

$^6C_4 = 15$

So, the table is as follows:

$\left[\begin{array}{ccc}TV&Good&Proportion\\1,2,3,4&2&0.5&2,3,4,5&2&0.5&3,4,5,6&2&0.5\\4,5,6,1&3&0.75&5,6,1,2&4&1&6,1,2,3&3&0.75\\1,2,3,5&3&0.75&3,5,6,2&3&0.75&1,3,4,5&2&0.5\\1,3,4,6&2&0.5&1,4,5,2&3&0.75&2,4,6,1&3&0.75\\2,4,6,3&2&0.5&2,4,6,5&3&0.75&3,5,6,1&3&0.75\end{array}\right]$

The proportion column is calculated by dividing the number of Good TVs by the total selected (4) i.e.

$p = \frac{Good}{n}$

<u>(d) The sampling distribution</u>

In (a), we have:

$p = 0.67$ --- proportion of Good TV

The sampling error is calculated as follows:

$SE_n = |p - p_n|$

So, we have:

$\left[\begin{array}{ccc}TV&Good&SE\\1,2,3,4&2&0.17&2,3,4,5&2&0.17&3,4,5,6&2&0.17\\4,5,6,1&3&0.08&5,6,1,2&4&0.33&6,1,2,3&3&0.08\\1,2,3,5&3&0.08&3,5,6,2&3&0.08&1,3,4,5&2&0.17\\1,3,4,6&2&0.17&1,4,5,2&3&0.08&2,4,6,1&3&0.08\\2,4,6,3&2&0.17&2,4,6,5&3&0.08&3,5,6,1&3&0.08\end{array}\right]$

Read more about sampling distributions at:

brainly.com/question/10554762

3 0

2 years ago