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Answer:
A)
For this distribution mean = np = 600*0.7 = 420
Standard Deviation = sqrt(npq) = aqrt(600*0.7*0.3) = 11.22
And the shape of the distribution is rightly skewed
B)
p = 0.7
n = 600
standard error, SE = sqrt(p(1-p)/n)
SE = sqrt(0.7(1-0.7) / 600) = 0.0187
Za/2 = Z0.025= 1.96
confidence interval:
p +/- [Za/2 * SE]
0.70 +/- 0.0367
( 0.6633 , 0.7367 )
lower bound = 0.6633
upper bound = 0.7367
p-hat is normally distributed with mean = p = 0.7, and standard deviation = 0.0187
c) In B. the shaded area shows probability 0.95 of that population proportion P lies in betweeen 0.663 and 0.737.
d) Here we need to find probability between 0.67 and 0.73
p ( 0.67 < P < 0.73 )
=
= p ( -1.604 < z < 1.604 )
= p ( Z < 1.604 ) - p ( z < -1.604 )
= 0.9456 - 0.0544
= 0.8913
E ) Here we need find exact probability ,
0.67 *600 = 402 , 0.73 *600 = 438
Here we have mean = 420 and standard deviation = 11.22
p ( 402 < x < 438 )
=
= p ( -1.604 < z < 1.604 )
= p ( Z < 1.604 ) - p ( z < -1.604 )
= 0.9456 - 0.0544
= 0.8913
F. at least 67%
p ( p 0.67 )
=
= p ( z -1.604)
= 1-p( z< -1.604)
=1-0.0544
= 0.9456
no more than 73%.
p ( p 0.73 )
=
= p ( z 1.604 )
= 0.9455
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