Respuesta :
Answer:
The conditions necessary for the analysis are met.
The probability the newspaper’s sample will lead them to predict defeat is 0.7881
Step-by-step explanation:
We are given;
population proportion; μ = 52% = 0.52
Sample size;n = 400
The conditions are;
10% conditon: sample size is less than 10% of the population size
Success or failure condition; np = 400 x 0.52 = 208 and n(1 - p) = 400(1 - 0.52) = 192.
Both values are greater than 10
Randomization condition; we assume that the voters were randomly selected.
So the conditions are met.
Now, the standard deviation is gotten from;
σ = √((p(1 - p)/n)
where;
p is the population proportion
n is the sample size
σ is standard deviation
Thus;
σ = √((0.52(1 - 0.52)/400)
σ = √((0.52(0.48)/400)
σ = 0.025
Now to find the z-value, we'll use;
P(p^ > 0.5) = P(z > (x - μ)/σ)
Thus;
P(p^ > 0.5) = P(z > (0.5 - 0.52)/0.025)
This gives;
P(p^ > 0.5) = P(z > - 0.8)
This gives;
P(p^ > 0.5) = 1 - P(z < -0.8)
From the table attached we have a z value of 0.21186
Thus;
P(p^ > 0.5) = 1 - 0.21186 = 0.7881
Thus, the probability the newspaper’s sample will lead them to predict defeat is 0.7871
Answer:
The probability, P that the newspaper's sample will lead them to predict defeat is 0.21186
Step-by-step explanation:
Here we have
p = 52%, therefore np = 208 Voters
q = 1 - p = 1 - 0.52 = 0.48
nq = 0.48×400 = 192 Voters
Therefore, as np and nq are both > 10 the conditions for approximation to normality are met
We therefore have
μ = 0.52 and
σ = [tex]\sqrt{\frac{pq}{n} } = \sqrt{\frac{0.52 \times 0.48}{400} } = 0.0249799 \approx 0.025[/tex]
The z score is therefore;
[tex]Z=\frac{x-\mu }{\sigma } =\frac{0.5-0.52 }{0.025 }= -0.8[/tex]
The probability the newspaper's sample will lead them to predict defeat is given by P(Z < -0.8) = 0.21186.
