Required probability is 0.8990
Given that,
p = 0.44
1 - p = 0.56
n = 736
[tex]\mu_\hat{p}} = p = 0.44\\\sigma_\hat{p}} = \frac{p*(1-p)}{n} = \sqrt(\frac{0.44*0.56)}{736} ) =[/tex]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<[tex]\hat{p}[/tex]<0.47) = P((0.41-0.44)/0.01830) < [tex]\frac{\hat{p} - \mu _{\hat{p}} }{\sigma _{\hat{p}} }[/tex]<(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:
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