Easy probability problem 4 Assume the sampling distribution

Easy probability problem !

4. Assume the sampling distribution of the sample proportion has mean p = 0.55. Based on samples of size n = 200. calculate the standard error of the sampling distribution of p. (1) 5. In problem 4, find (1) a. P(0.50

Solution

Q4.
Standard Error = Sqrt(p*(1-p)/n))
x = Mean
n = Sample Size,
Mean(x)=55
Sample Size(n)=100
Sample proportion =0.55
Standard Error = Sqrt ( (0.55*0.45) /100) )
= 0.0497

Q5.
Proportion ( P ) =0.55
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.55*0.45/200)= P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.96
P( x-u/s.d < x - 0.55/0.03517 ) = 0.025
That is, ( x - 0.55/0.03517 ) = -1.96
--> x = -1.96 * 0.03517 + 0.55 = 0.4811                  

Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.5) = (0.5-0.55)/0.0352
= -0.05/0.0352 = -1.4205
= P ( Z <-1.4205) From Standard Normal Table
= 0.07774
P(X < 0.6) = (0.6-0.55)/0.0352
= 0.05/0.0352 = 1.4205
= P ( Z <1.4205) From Standard Normal Table
= 0.92226
P(0.5 < X < 0.6) = 0.92226-0.07774 = 0.8445                  

b)
P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.96
P( x-u/s.d < x - 0.55/0.03517 ) = 0.025
That is, ( x - 0.55/0.03517 ) = -1.96
--> x = -1.96 * 0.03517 + 0.55 = 0.4811