70 numbers are rounded off to the nearest integer and then summed. if the individual round-off error are uniformly distributed over (−.5,.5) what is the probability that the resultant sum differs from the exact sum by more than 4?

0.147

Step-by-step explanation:

NB: check exclusive solution on the file attached.

This is a limit theorem problem. To solve this problem, we have to make series of symbol assumptions.

We say that Xi is equal to the ith round off error. Therefore, Xi~ UNIF (−0.5,0.5). Also, the f(Xi)= 1( that is the density). When Xi is an element of -0.5,0.5; i=1,...,70.

Step one(1): calculate the E(Xi). And, in order to calculate E(Xi), we use;

E(Xi) = 0.5+(-0.5)/2

E(Xi)= 0.

Step two(2) : calculate the var(Xi). Which is; 0.5-(0.5^2)/12 = 1/12.

X(1) to X(70) are independent. Therefore, using the central limit theorem we know approximately that normal distribution is U= 50∗0=0. Also, σ^2= 70 × 1/12 = 5.83.

Step three(3): we find our P from the Z values(z values can be gotten from the z-standard deviation from cut off.

2p(Z>1.66)

= .147

i would pic false cause it goes down towards negative

step-by-step explanation:

i think the answer is fried chicken ; >

step-by-step explanation: