Question

A random sample of size n = 60 is selected from a binomial distribution with population proportion p = 0.25.

Answer 1

Answer:

a) $p \sim N (p, \sqrt{\frac{p(1-p)}{n}})$

With the following parameters:

$\mu_p = 0.25$

$\sigma_p = \sqrt{\frac{0.25*(1-0.25)}{60}}= 0.0559$

b) $\mu_p = 0.25$

$\sigma_p = \sqrt{\frac{0.25*(1-0.25)}{60}}= 0.0559$

c) $P(\frac{0.13-0.25}{0.0559}< Z< \frac{0.47-0.25}{0.0559}) = P(-2.147< Z< 3.936)$

And for this case we can use the following difference and the normal standard distribution table or excel and we got:

$P(-2.147< Z< 3.936) = P(z$

Step-by-step explanation:

For this case we have the following info:

n = 60 represent the sample size

p = 0.25 represent the proportion of success

For this case we can check the conditions to use the normal distribution:

1) np= 60*0.25=15>10

2) n(1-p)= 60(1-0.25) = 45>10

So then we can use the normal distribution as an approximate distribution for p

Part a

$p \sim N (p, \sqrt{\frac{p(1-p)}{n}})$

Part b

With the following parameters:

$\mu_p = 0.25$

$\sigma_p = \sqrt{\frac{0.25*(1-0.25)}{60}}= 0.0559$

Part c

For this case we want this probability:

$P(0.13< p$

And for this case we can use the z score given by:

$z = \frac{p -\mu_p}{\sigma_p}$

And using this formula we got:

$P(\frac{0.13-0.25}{0.0559}< Z< \frac{0.47-0.25}{0.0559}) = P(-2.147< Z< 3.936)$

And for this case we can use the following difference and the normal standard distribution table or excel and we got:

$P(-2.147< Z< 3.936) = P(z$