Answer:
21.21%
Step-by-step explanation:
Assume that proportion of support follows normal distribution.For resulting in defeat the proportion of support should less than 0.5.Given that the proportion of support is p= 0.52.
Sample size(n)= 402
We know that standard deviation is given as
![\sigma = \sqrt {\dfrac {p(1-p)}{n}}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%20%7B%5Cdfrac%20%7Bp%281-p%29%7D%7Bn%7D%7D)
Now by putting the values
![\sigma = \sqrt {\dfrac{0.52\times 0.48}{402}}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%20%7B%5Cdfrac%7B0.52%5Ctimes%200.48%7D%7B402%7D%7D)
σ=0.0249
Proportion of support follows normal distribution with mean is 0.52 and standard deviation is 0.0249
We know that ![Z=\dfrac{\bar{X}-\mu }{\sigma }](https://tex.z-dn.net/?f=Z%3D%5Cdfrac%7B%5Cbar%7BX%7D-%5Cmu%20%7D%7B%5Csigma%20%7D)
So
![Z=\dfrac{0.5-0.52 }{0.0249}](https://tex.z-dn.net/?f=Z%3D%5Cdfrac%7B0.5-0.52%20%7D%7B0.0249%7D)
Z= -0.80
So P(Z<-.80) =0.2118 from standard table.
So the probability of news paper to predict defeat is 21.21%
Cid Sam is a way to remember
C -control S - SAME (keep the same)
I -independent A- alter
D-depend M- measure
independent- number of downloads
Dependant - money spent
Answer:
sHOULD BE 7:2
Step-by-step explanation: