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svp [43]
3 years ago
14

An instructor of a large college class gave an exam that has a possible total of 100 points. The instructor records the scores o

f 100 students from his class and produced the following histogram. The instructor says any score above a 90 is an A. Scores in the 80 to 89 range would be a B, scores between 70 and 79 would yield a C and below 70 would result in a failing grade.Which of the following are true?Select all that apply.a.The median grade lies between 76 and 84.b.The median grade is greater than the mean grade.c.The median grade is an Ad.The median grade is a B.e.The mean grade is greater than the median grade.f.The mean grade and the median grade are equal.
Mathematics
1 answer:
pishuonlain [190]3 years ago
4 0

Answer:

(a) the median grade lies between 76 and 84

(b) the Median grade is greater than the mean grade.

(d) the median grade is a B

Step-by-step explanation:

consider the histogram carefully and answer.

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Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute in
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Answer:

a) Null hypothesis:p\geq 0.53  

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b) z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429  

p_v =P(Z

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Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

\hat p=0.46 estimated proportion of American families owning stocks or stock funds

p_o=0.53 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

Part a

We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:  

Null hypothesis:p\geq 0.53  

Alternative hypothesis:p < 0.53  

Part b

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429  

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The significance level provided \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z

Part c  

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