Question

Suppose 44% of the students in a university are baseball players. If a sample of 736 students is selected, what is the probability that the sample proportion of baseball players will differ from the population proportion by less than 3%

Answer 1

Required probability is 0.8990

Given that,

p = 0.44

1 - p = 0.56

n = 736

$\mu_\hat{p}} = p = 0.44\\\sigma_\hat{p}} = \frac{p*(1-p)}{n} = \sqrt(\frac{0.44*0.56)}{736} ) =$0.01830

Converting Normal distribution to standard normal because  A normal distribution is determined by two parameters the mean and the variance. A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution. Hence it's easy to work with standard normal distribution.

P(0.14<$\hat{p}$<0.47) =  P((0.41-0.44)/0.01830) < $\frac{\hat{p} - \mu _{\hat{p}} }{\sigma _{\hat{p}} }$<(0.47-0.44)/0.01830

Using normal algebra we get:

= P(-1.64 < z < 1.64)

= P(z < 1.64) - P(z < -1.64)

= 0.9495 - 0.0505

= 0.8990

Thus, required probability is 0.8990

To learn more about normal distribution visit:

brainly.com/question/4079902

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