Question

Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 products randomly from the line each hour. Looking at the binomial distribution, if fewer than how many are within specifications would require that the production line be shut down (unusual) and repaired?Fewer than 10Fewer than 12Fewer than 11Fewer than 9

Answer 1

Answer:

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

Step-by-step explanation:

There is a random binomial variable $X$ that represents the number of units come off the line within product specifications in a review of $n$ Bernoulli-type trials with probability of success $0.91$. Therefore, the model is ${15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}$. So:

$P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002$

$P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013$

$P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082$

$P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399$

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399