Answer:
The probability that in a given day the store receives four or less bad checks is 0.70.
The probability that in a given day the store receives more than 3 bad checks is 0.50.
Step-by-step explanation:
The data provided shows the number of bad checks received by the management of a grocery store for a period of 200 days.
The probability distribution and the cumulative probability distribution are shown in the table attached below.
Let the number of bad checks received in a day be represented by <em>X</em>.
Compute the probability that in a given day the store receives four or less bad checks as follows:
P (X ≤ 4) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

Thus, the probability that in a given day the store receives four or less bad checks is 0.70.
Compute the probability that in a given day the store receives more than 3 bad checks as follows:
P (X > 3) = 1 - P (X ≤ 3)
= 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)]
![=1-[0.04+0.06+0.10+0.30]\\=1-0.50\\=0.50](https://tex.z-dn.net/?f=%3D1-%5B0.04%2B0.06%2B0.10%2B0.30%5D%5C%5C%3D1-0.50%5C%5C%3D0.50)
Thus, the probability that in a given day the store receives more than 3 bad checks is 0.50.