Answer: 1 < x < 2
Step-by-step explanation:
Actually, distributive property isn't needed to combine this. You can simply subtract 7 from 13 which will give you 6. The solution is 6c.
Let
![X](https://tex.z-dn.net/?f=X)
denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by
![X_1,\ldots,X_{36}](https://tex.z-dn.net/?f=X_1%2C%5Cldots%2CX_%7B36%7D)
, each independently and identically distributed with distribution
![X_i\sim\mathcal N(26,7.2)](https://tex.z-dn.net/?f=X_i%5Csim%5Cmathcal%20N%2826%2C7.2%29)
.
You want to find
![\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)](https://tex.z-dn.net/?f=%5Cmathbb%20P%28X_1%2B%5Ccdots%2BX_%7B36%7D%3E1000%29%3D%5Cmathbb%20P%5Cleft%28%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E%7B36%7DX_i%3E1000%5Cright%29)
Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to
![\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)](https://tex.z-dn.net/?f=%5Cmathbb%20P%5Cleft%2836%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E%7B36%7D%5Cfrac%7BX_i%7D%7B36%7D%3E1000%5Cright%29%3D%5Cmathbb%20P%5Cleft%28%5Coverline%20X%3E%5Cdfrac%7B1000%7D%7B36%7D%5Cright%29)
Recall that if
![X\sim\mathcal N(\mu,\sigma)](https://tex.z-dn.net/?f=X%5Csim%5Cmathcal%20N%28%5Cmu%2C%5Csigma%29)
, then the sampling distribution
![\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)](https://tex.z-dn.net/?f=%5Coverline%20X%3D%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5En%5Cfrac%7BX_i%7Dn%5Csim%5Cmathcal%20N%5Cleft%28%5Cmu%2C%5Cdfrac%5Csigma%7B%5Csqrt%20n%7D%5Cright%29)
with
![n](https://tex.z-dn.net/?f=n)
being the size of the sample.
Transforming to the standard normal distribution, you have
![Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}](https://tex.z-dn.net/?f=Z%3D%5Cdfrac%7B%5Coverline%20X-%5Cmu_%7B%5Coverline%20X%7D%7D%7B%5Csigma_%7B%5Coverline%20X%7D%7D%3D%5Csqrt%20n%5Cdfrac%7B%5Coverline%20X-%5Cmu%7D%7B%5Csigma%7D)
so that in this case,
![Z=6\dfrac{\overline X-26}{7.2}](https://tex.z-dn.net/?f=Z%3D6%5Cdfrac%7B%5Coverline%20X-26%7D%7B7.2%7D)
and the probability is equivalent to
![\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)](https://tex.z-dn.net/?f=%5Cmathbb%20P%5Cleft%28%5Coverline%20X%3E%5Cdfrac%7B1000%7D%7B36%7D%5Cright%29%3D%5Cmathbb%20P%5Cleft%286%5Cdfrac%7B%5Coverline%20X-26%7D%7B7.2%7D%3E6%5Cdfrac%7B%5Cfrac%7B1000%7D%7B36%7D-26%7D%7B7.2%7D%5Cright%29)
The answer is 95%
This is something you should memorize. Specifically it is from the Empirical Rule (or 68-95-99.7 rule) which gives rough approximations of areas under the curve.
Answer: 42 pop tunes
Step-by-step explanation:
35/5=7
6*7=42