Respuesta :
Answer:
Probability that 32 or more from this sample used Internet Explorer as their browser is 0.9015.
Step-by-step explanation:
We are given that according to Net Market Share, Microsoft's Internet Explorer browser has 53.4% of the global market.
A random sample of 70 users was selected.
Let [tex]\hat p[/tex] = sample proportion of users who used Internet Explorer as their browser.
The z score probability distribution for sample proportion is given by;
Z = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, p = population proportion of users who use internet explorer = 53.4%
[tex]\hat p[/tex] = sample proportion = [tex]\frac{32}{70}[/tex] = 0.457
n = sample of users = 70
Now, probability that 32 or more from this sample used Internet Explorer as their browser is given by = P( [tex]\hat p[/tex] [tex]\geq[/tex] 0.457)
P( [tex]\hat p[/tex] [tex]\geq[/tex] 0.457) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] [tex]\geq[/tex] [tex]\frac{0.457-0.534}{\sqrt{\frac{0.457(1-0.457)}{70} } }[/tex] ) = P(Z [tex]\geq[/tex] -1.29)
= P(Z [tex]\leq[/tex] 1.29) = 0.9015
The above probability is calculated by looking at the value of x = 1.29 in the z table which has an area of 0.9015.
