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notka56 [123]
2 years ago
13

A poll shows that 50% of students play sports

Mathematics
1 answer:
Firlakuza [10]2 years ago
8 0

Answer:

The chance of getting a sample  proportion of 70% or greater is 0.026.

Step-by-step explanation:

We are given that a poll shows that 50% of students play sports .

A random sample of 20 students showed that  70% of them play sports.

Let \hat p = sample proportion of students who play sports

The z-score probability distribution for the sample proportion is given by;

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

where, \hat p = sample proportion of students who play sports = 70%

           p = population proportion of students who play sports = 50%

           n = sample of students = 20

Now, the chance of getting a sample  proportion of 70% or greater is given by = P(\hat p \geq 70%)

   P(\hat p \geq 70%) = P( \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } \geq \frac{0.70-0.50}{\sqrt{\frac{0.70(1-0.70)}{20} } } ) = P(Z \geq 1.95) = 1 - P(Z < 1.95)

                                                                 = 1 - 0.97441 = <u>0.026</u>

The above probability is calculated by looking at the value of x = 1.95 in the z-table which has an area of 0.97441.

Hence, the chance of getting a sample  proportion of 70% or greater is 0.026.

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Step-by-step explanation:

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Answer:

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It is not D, I just had this question.
serg [7]
Oh ok cool, thanks for the extra points though
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