This is all I got sorry if it don’t help
Answer:
happy birthday may god bless you
F = 1.8F + 32
F - 1.8F = 1.8F - 1.8F + 32
-0.8F = 32
-0.8F / -0.8= 32 / -0.8
F = -40
I have an expression
![\sigma = \sqrt{p(1-p)/n}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7Bp%281-p%29%2Fn%7D)
floating around in my head; let's see if it makes sense.
The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is
![E( (b-p)^2 ) = (-p)^2(1-p) + (1-p)^2p = p(1-p)](https://tex.z-dn.net/?f=E%28%20%28b-p%29%5E2%20%29%20%3D%20%20%28-p%29%5E2%281-p%29%20%2B%20%281-p%29%5E2p%20%3D%20p%281-p%29)
That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:
![\sigma = \sqrt{p(1-p)/n}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7Bp%281-p%29%2Fn%7D)
Plugging in the numbers,
![\sigma = \sqrt{.24(1-.24)/490} = 0.019 = 1.9\%](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B.24%281-.24%29%2F490%7D%20%3D%200.019%20%3D%201.9%5C%25)
One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.
So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.