Do Independent Normal Variables Exceed a Given Total?

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To determine the probability that the total of seven independent random variables from a normal distribution N(10, 7) exceeds 88, one must first calculate the mean and variance of the sum. The mean of the total is 70, while the variance is 49, leading to a standard deviation of 7. Standardizing the total using the formula (x - μ) / σ is crucial, and the correct approach involves using the standard deviation rather than the variance. After proper normalization, the probability can be computed, which should yield the expected result of 0.005. Understanding these statistical principles is essential for accurate calculations in probability.
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Hi
I'm wondering if someone can help me out on this question as to how to go about doing it:
X_1, X_2... X_7 are independent random variables represnting a random sample of size 7 from the normal N(10, 7) distribution. Find to 3 dp probablitity that the sample total exceeds 88.

I tried to standardise this but my numbers don't seem to get me the answer of 0.005. Can someone help me out? Thanks in advance :)
 
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How did try to do it? Remember, the d.f. of the sum of random variables with normal distributions is another normal distribution with a mean that is the sum of the means of the individual variables and a variance that is the sum of the variances of the individual variables. Also remember when changing varibles that what appears in the the normal distribution is \frac{(x-\mu)}{\sigma} and not \frac{(x-\mu)}{\sigma^2}, so use the standard deviation and not the variance when changing variables.
 
Thanks for the help! when i originally did it i used (88-10)/7 to try and standardise it - giving me a ridiculous number. So with the normal distribution N(10,7) what should I now do with those - I'm still kind of confused...
 
Remember, the mean is the sum of the means of the X_i, so that's 10+10+10+...=70. The variance is the sum of the variances. Remember to normalize with the standard deviation and not the variance. Once you do that you do get the answer you said you were supposed to.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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