Question:
Suppose a simple random sample of size n = 1000 is obtained from a population whose size is N = 1,000,000 and whose population proportion with a specified characteristic is p = 0.44 . Complete parts​ (a) through​ (c) below.
​(a) Describe the sampling distribution of ModifyingAbove p with caret.
A.)Approximately​ normal, mu Subscript ModifyingAbove p with caretequals0.44 and sigma Subscript ModifyingAbove p with caretalmost equals0.0002
B.)Approximately​ normal, mu Subscript ModifyingAbove p with caretequals0.44 and sigma Subscript ModifyingAbove p with caretalmost equals0.0005
C.)Approximately​ normal, mu Subscript ModifyingAbove p with caretequals0.44 and sigma Subscript ModifyingAbove p with caretalmost equals0.0157
​(b) What is the probability of obtaining xequals480 or more individuals with the​ characteristic?
​P(xgreater than or equals480​)equals
nothing ​(Round to four decimal places as​ needed.)
​(c) What is the probability of obtaining xequals410 or fewer individuals with the​ characteristic?
​P(xless than or equals410​)equals
nothing ​(Round to four decimal places as​ needed.)
Answer:
a) Option C.
b) 0.1021
c) 0.0280
Step-by-step explanation:
Given:
Sample size, n = 1000
p' = 0.44
a) up' = p' = 0.44
The sampling distribution will be:
[tex] \sigma _p' = \sqrt{\frac{p' (1 - p')}{n}}[/tex]
[tex] = \sqrt{\frac{0.44 (1 - 0.44)}{1000}}[/tex]
[tex] = \sqrt{\frac{0.44*0.56}{1000}} = 0.0157 [/tex]
Option C is correct.
b) The probability when x ≥ 460
[tex] P' = \frac{x}{n} = \frac{460}{1000} = 0.46[/tex]
p'(P ≥ 0.46)
[tex] 1 - P = \frac{(p' - up')}{\sigma _p'} < \frac{0.46 - 0.44}{0.0157} [/tex]
[tex] = 1-P( Z < 1.27) [/tex]
From the normal distribution table
NORMSDIST(1.27) = 0.898
1-0.8979 = 0.1021
Therefore, the probability = 0.102
c) x ≤ 410
[tex] P' = \frac{x}{n} = \frac{410}{1000} = 0.41[/tex]
p'(P ≤ 0.41)
[tex] P = \frac{(p' - up')}{\sigma _p'} < \frac{0.41 - 0.44}{0.0157} [/tex]
[tex] = P( Z < - 1.9108) [/tex]
From the normal distribution table
NORMSDIST(-1.9108) = 0.0280
Probability = 0.0280
