Answer:
Step-by-step explanation:
Hello!
The Central Limit Theorem allows you to approximate the sampling distribution of a variable to normal. In this case, if n≥30, you can approximate the distribution of the sample proportion to normal:
^p≈N(n*p; [p(1-p)]/n)
You need to calculate the percentage of the distribution that is between the two values of the sample proportion of 0.50 and 0.60, symbolically:
P(0.50≤^p≤0.60)
P(^p≤0.60)-P(^p≤0.50)
To reach the value of accumulated probabilities until 0.6 and 0.5 you have to standardize both values by subtracting the mean of the distribution (np) and dividing it by the standard deviation (√ [p(1-p)]/n)
Z=[tex]\frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]
^p= 0.6⇒ [tex]Z= \frac{0.6-0.54}{\sqrt{\frac{0.54*0.46}{111} } } = 1.27[/tex]
^p= 0.5 ⇒ [tex]Z= \frac{0.5-0.54}{\sqrt{\frac{0.54*0.46}{111} } } = -0.85[/tex]
Then:
P(Z≤1.27) - P(Z≤-0.85)= 0.89796 - 0.19766= 0.7003
I hope this helps!
