Question

TubeView, a digital television distribution company based in Gatineau (QC), hires a marketing firm to gauge the satisfaction levels of their customers. The CEO of TubeView has claimed that past polls have suggested that the customer satisfaction level is at 90%.
If this claim was true, what is the probability that in a random sample of 12 customers:
a) Exactly 8 customers are satisfied with TubeView’s service?
b) At least 8 customers are satisfied with TubeView’s service?
c) All of the customers they contact are satisfied with TubeView’s service?
d) At most 2 customers are not satisfied with TubeView’s service?
e) P(5 ≤ X < 9) where X = satisfied customer?

Answer 1

Answer on Question #60349 – Math – Statistics and Probability

Question

TubeView, a digital television distribution company based in Gatineau (QC), hires a marketing firm to gauge the satisfaction levels of their customers. The CEO of TubeView has claimed that past polls have suggested that the customer satisfaction level is at 90%.

If this claim was true, what is the probability that in a random sample of 12 customers:

a) Exactly 8 customers are satisfied with TubeView’s service?

b) At least 8 customers are satisfied with TubeView’s service?

c) All of the customers they contact are satisfied with TubeView’s service?

d) At most 2 customers are not satisfied with TubeView’s service?

e) $P(5 \leq X < 9)$ where $X =$ satisfied customer?

Solution

Verifying conditions required to apply the normal approximation for the binomial distribution:

$$n p = 12 \times 0.9 = 10.8 > 5 \text{ passed}$$ $$n q = 12 \times 0.1 = 1.2 < 5 \text{ failed}$$

Therefore, one cannot use the normal approximation for the binomial distribution.

The probabilities can be calculated using the binomial formula:

$$P(x) = {}_n C_x p^x q^{n-x} = \frac{n!}{(n-x)! x!} p^x q^{n-x}$$

a) $P(x = 8) = \frac{12!}{(12 - 8)!8!} \times 0.9^8 \times 0.1^{12 - 8} = 0.0213$

b) $P(x \geq 8) = P(x = 8) + P(x = 9) + P(x = 10) + P(x = 11) + P(x = 12)$

$$P(x = 9) = \frac{12!}{(12 - 9)!9!} \times 0.9^9 \times 0.1^{12 - 9} = 0.0852$$ $$P(x = 10) = \frac{12!}{(12 - 10)!10!} \times 0.9^{10} \times 0.1^{12 - 10} = 0.2301$$ $$P(x = 11) = \frac{12!}{(12 - 11)!11!} \times 0.9^{11} \times 0.1^{12 - 11} = 0.3766$$ $$P(x = 12) = \frac{12!}{(12 - 12)!12!} \times 0.9^{12} \times 0.1^{12 - 12} = 0.2824$$ $$P(x \geq 8) = 0.0213 + 0.0852 + 0.2301 + 0.3766 + 0.2824 = 0.9956$$

c) $P(x = 12) = \frac{12!}{(12 - 12)!12!} \times 0.9^{12} \times 0.1^{12 - 12} = 0.2824$

d) The event "at most 2 customers are not satisfied" is equal to "at least 10 customers satisfied"

$$P(x \geq 10) = P(x = 10) + P(x = 11) + P(x = 12)$$ $$P(x \geq 10) = 0.2301 + 0.3766 + 0.2824 = 0.8891$$

e) $P(5 \leq x < 9) = P(x = 5) + P(x = 6) + P(x = 7) + P(x = 8)$

$$P(x = 5) = \frac{12!}{(12 - 5)!5!} \times 0.9^{5} \times 0.1^{12 - 5} = 0$$ $$P(x = 6) = \frac{12!}{(12 - 6)!6!} \times 0.9^{6} \times 0.1^{12 - 6} = 0.0005$$ $$P(x = 7) = \frac{12!}{(12 - 7)!7!} \times 0.9^{7} \times 0.1^{12 - 7} = 0.0038$$ $$P(x = 8) = \frac{12!}{(12 - 8)!8!} \times 0.9^{8} \times 0.1^{12 - 8} = 0.0213$$ $$P(5 \leq x < 9) = 0 + 0.0005 + 0.0038 + 0.0213 = 0.0256.$$

Answer:

a) 0.0213;

b) 0.9956;

c) 0.2824;

d) 0.8891;

e) 0.0256.

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