Find the quotient: 7/8 ÷ 3/8 A 21/64 B 3/7 C 2 1/3 D 7 1/3

The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q

Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

$\frac{\sum f}{4}=\frac{100}{4}=25$

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF) $_{p}$ = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

$=34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75$

The value of first quartile is 34.75.

Q₃ is at the position:

$\frac{3\sum f}{4}=\frac{3\times100}{4}=75$

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF) $_{p}$ = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

$=44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83$

The value of third quartile is 44.83.

Then the inter-quartile range is:

$IQR = Q_{3}-Q_{1}$

$=44.83-34.75\\=10.08$

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

$P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

$=\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89$

Thus, the probability that a student scored at most 50 on the examination is 0.89.

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On the school playground, the slide is 8 feet due west of the tire swing and 15 feet due south of the monkey bars. What is the d

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7 0

3 years ago

Read 2 more answers

A large restaurant is being sued for age discrimination because 15% of newly hired candidates are between the ages of 30 years a

lesya [120]

Answer:

Part A:

The null and alternative hypothesis are:

$H_0: \pi=0.5\\\\H_1: \pi\neq 0.5$

Part B:

- A Type I error is when the null hypothesis is rejected although it is true. In this case, it would mean that we conclude that the hiring process is discriminatory, when in reality it is a random result and the process is not discriminatory.

- A Type II error is when the null hypothesis fails to be rejected although it is false. In this case, the hiring process is discriminatory, but statistically the result is not significant enough to prove that.

Part C:

A reduction in the significance level causes a reduction in the power of the test.

Part D:

The power of the test is increased with a larger sample.

Step-by-step explanation:

We have a restaurant with hire a proprotion of 15 % of people in the age ragne of 30-50 years. The expected proportion, according to the applicants, is 50%.

The test will tell us if the actual 15% is a result of a discriminatory practice or a random result.

Part A:

The null and alternative hypothesis are:

$H_0: \pi=0.5\\\\H_1: \pi\neq 0.5$

Part B:

- A Type I error is when the null hypothesis is rejected although it is true. In this case, it would mean that we conclude that the hiring process is discriminatory, when in reality it is a random result and the process is not discriminatory.

- A Type II error is when the null hypothesis fails to be rejected although it is false. In this case, the hiring process is discriminatory, but statistically the result is not significant enough to prove that.

Part C:

The power of an hypothesis test is the probability that the test rejects the null hypothesis (H_0) when a specific alternative hypothesis (H_1) is true.

If the significance level is reduced (from 5% to 1%), the rejection region is reduced, so the probability of rejecting the null hypothesis is also reduced.

Then, a reduction in the significance level causes a reduction in the power of the test.

Part D:

A bigger sample size gives robustness to the sample statistic. Then, if the alternative hypothesis is true, the probabilities of detecting the effect are increased with increased sample size.

Then, the power of the test is increased with a larger sample.

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What is the least natural number that is greater than the largest 4-?digit numbers

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Find the value of w.

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Answer:

w = 3.2 m