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lyudmila [28]
3 years ago
13

The formula for figuring a​ student's average test score A is A equals = StartFraction p Over t EndFraction p t ​, where p is th

e total points earned on all the​ tests, and t is the number of tests. Find a​ student's average when 345 points are earned on 5 tests.
Mathematics
1 answer:
RideAnS [48]3 years ago
7 0

Answer:

Average = 69.

Step-by-step explanation:

A = p /t

When p =345 and t = 5

A = 345 /5

= 69.

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One True Love? A survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each pe
sertanlavr [38]

Answer:

99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

Step-by-step explanation:

We are given that a survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each person." has been conducted. The result is that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered ‘‘don’t know."

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

                                 P.Q. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of people who disagree with the statement = \frac{1812}{2625} = 0.69

           n = sample of respondents = 2625

           p = population proportion of people who disagree with statement

<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>

<u>So, 99% confidence interval for the population proportion, p is ;</u>

P(-2.58 < N(0,1) < 2.58) = 0.99  {As the critical value of z at 0.5% level

                                                   of significance are -2.58 & 2.58}  

P(-2.58 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.58) = 0.99

P( -2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<u>99% confidence interval for p</u> = [ \hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

= [ 0.69-2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } } , 0.69+2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } } ]

 = [0.667 , 0.713]

Therefore, 99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

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