Assume that 26% of students at a university wear contact lenses. A teacher randomly picks 300 students. Describe the sampling distribution model of the proportion of students in this group who wear contact lenses. Round to the nearest tenth.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]

Solution to the problem

For this case we can conclude that the population proportion present the following distribution:

[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]

And the reason is because np>10 and n(1-p)>10

[tex] \mu_p = p = 0.26[/tex]

And the standard error is given by:

[tex]\sigma_p = \sqrt{\frac{p(1-p)}{n}}[/tex]

And replacing we got:

[tex]\sigma_p = \sqrt{\frac{0.26(1-0.26)}{300}}= 0.025[/tex]

Assume that​ 26% of students at a university wear contact lenses. A teacher randomly picks 300 students. Describe the sampling distribution model of the proportion of students in this group who wear contact lenses. Round to the nearest tenth.

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Assume that 26% of students at a university wear contact lenses. A teacher randomly picks 300 students. Describe the sampling distribution model of the proportion of students in this group who wear contact lenses. Round to the nearest tenth.