Question

1. The management of a grocery store has kept a record of bad checks received per day for a period of 300 days. The data are shown below.

Number of Bad
Checks Received Number of Days
0 12
1 15
2 35
3 55
4 63
5 35
6 38
7 32
8 15

a) Develop a probability distribution for the above data.
b) Is the probability distribution that you found in Part “a” a proper probability distribution? Explain.
c) Determine the cumulative probability distribution, F(x).
d) What is the probability that in a given day the store receives four or less bad checks?
e) What is the probability that in a given day the store receives more than two bad checks?
f) What is the expected value of the number of checks received?
g) Compute the variance of the number of checks received.
h) Compute the standard deviation of number of checks received.

 

Answer 1

Answer on Question #76845 – Math – Statistics and Probability

The management of a grocery store has kept a record of bad checks received per day for a period of 300 days.

Question

a) Develop a probability distribution for the above data.

Solution

Question

b) Is the probability distribution that you found in Part “a” a proper probability distribution? Explain.

Solution

This is a proper probability distribution since:

$$p(x_i) \geq 0 \text{ for all } x$$ $$\sum_{i=0}^{8} p(x_i) = 1$$

**Question**

c) Determine the cumulative probability distribution, $F(x)$.

**Solution**

$$F(x) = \sum_{i: x_i \leq x} p_i$$ $$F(0) = p(x = 0) = 0.04$$ $$F(1) = p(x \leq 1) = 0.04 + 0.05 = 0.09$$ $$F(2) = p(x \leq 2) = 0.04 + 0.05 + 0.117 = 0.27$$ $$F(3) = p(x \leq 3) = 0.04 + 0.05 + 0.117 + 0.183 = 0.39$$ $$F(4) = p(x \leq 4) = 0.39 + 0.21 = 0.6$$ $$F(5) = p(x \leq 5) = 0.6 + 0.117 = 0.717$$ $$F(6) = p(x \leq 6) = 0.717 + 0.126 = 0.843$$ $$F(7) = p(x \leq 7) = 0.843 + 0.107 = 0.95$$ $$F(8) = p(x \leq 8) = 1$$

**Question**

d) What is the probability that in a given day the store receives four or less bad checks?

**Solution**

$$p(x \leq 4) = F(4) = 0.21 + 0.183 + 0.117 + 0.05 + 0.04 = 0.6$$

Question

e) What is the probability that in a given day the store receives more than two bad checks?

Solution

$$p(x > 2) = 1 - p(x \leq 2) = 1 - 0.04 - 0.05 - 0.117 = 0.793$$

Question

f) What is the expected value of the number of checks received?

Solution

$$\mu = \sum_{i=0}^{8} p_i x_i = 0.05 + 0.117 \cdot 2 + 0.183 \cdot 3 + 0.21 \cdot 4 + 0.117 \cdot 5 + 0.126 \cdot 6 + 0.107 \cdot 7 + 0.05 \cdot 8 = 4.163$$

Question

g) Compute the variance of the number of checks received.

Solution

$$var(x) = \sum_{i=0}^{8} p_i x_i^2 - \mu^2 = 0.05 + 0.117 \cdot 4 + 0.183 \cdot 9 + 0.21 \cdot 16 + 0.117 \cdot 25 + 0.126 \cdot 36 + 0.107 \cdot 49 + 0.05 \cdot 64 - 4.163^2 = 4.098$$

Question

h) Compute the standard deviation of number of checks received.

Solution

$$\sigma = \sqrt{var(x)} = \sqrt{4.098} = 2.024$$

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