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Magazine Chapter 6: Problem 50
Suppose 1.5 percent of the antennas on new Nokia cell phones are defective. For a random sample of 200 antennas, find the probability that: a. None of the antennas is defective. b. Three or more of the antennas are defective.
Short Answer
Expert verified
a. 0.049
b. 0.578
Step by step solution
01
Identify Known Variables
We know that the probability of a defective antenna is 1.5%, or 0.015. The number of antennas we are considering is 200.
02
Use Binomial Formula for Probability
Since each antenna can either be defective or not, this is a binomial distribution problem. The probability of exactly \( k \) defective antennas out of \( n \) is given by \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( p = 0.015 \) and \( n = 200 \).
03
Calculate Probability of Zero Defective Antennas
For part a, we want \( P(X = 0) \). Using the formula, this is \( \binom{200}{0} (0.015)^0 (1-0.015)^{200} \). Simplify to \( P(X = 0) = (1-0.015)^{200} \).
04
Simplify and Calculate P(X = 0)
Compute \( (1-0.015)^{200} \). This results in approximately \( 0.049 \).
05
Identify Complement Probability for Three or More
For part b, calculate the probability that three or more antennas are defective by using the complement rule: \( P(X \geq 3) = 1 - P(X < 3) \).
06
Compute P(X < 3)
\( P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \).
07
Calculate P(X = 1)
\( P(X = 1) = \binom{200}{1} (0.015)^1 (1-0.015)^{199} = 200 \times 0.015 \times (1-0.015)^{199} \). Compute the value, which is approximately \( 0.148 \).
08
Calculate P(X = 2)
\( P(X = 2) = \binom{200}{2} (0.015)^2 (1-0.015)^{198} \). Calculate to find approximately \( 0.225 \).
09
Sum Probabilities for P(X < 3)
Add \( P(X = 0) + P(X = 1) + P(X = 2) \). This is approximately \( 0.049 + 0.148 + 0.225 = 0.422 \).
10
Calculate P(X 鈮� 3)
Finally, \( P(X \geq 3) = 1 - P(X < 3) = 1 - 0.422 = 0.578 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Probability in Binomial Distribution
Probability is a fundamental concept in statistics that represents the likelihood of an event occurring. When dealing with binomial distribution, as in the case of defective antennas, the probability remains constant for each trial. It describes the chances of a singular outcome from two possible outcomes, such as an antenna being defective or not defective.
In a binomial distribution:
In a binomial distribution:
- The total number of trials is fixed, often represented by \( n \).
- Each trial is independent, meaning the outcome of one trial does not affect the outcome of another.
- The probability \( p \) of success (e.g., finding a defective antenna) is the same for each trial.
- The distribution is defined by two parameters: \( n \) (number of trials) and \( p \) (probability of success).
Utilizing the Complement Rule
The complement rule is a powerful tool in probability, often used when calculating the likelihood of 'at least' or 'at most' events. In the context of our exercise, to find the probability that three or more antennas are defective, it's much easier to first find the probability of less than three and then subtract it from one.
The complement rule essentially states:
The complement rule essentially states:
- For an event \( A \), the probability of \( A \) not occurring is \( 1 - P(A) \).
- If you calculate \( P(X < 3) \), which means zero, one, or two defective antennas, then \( P(X \geq 3) \) can be easily computed as \( 1 - P(X < 3) \).
Identifying Defective Products
In many industrial or manufacturing processes, the likelihood of defective products is a crucial measure for quality control. When we refer to defective products in probabilistic terms, it often operates under the assumption that each unit (or product, like an antenna) has a fixed probability of being defective.
Here, with a probability of 1.5% per antenna being defective, quality assurance teams can plan for expected levels of defects. This helps in:
Here, with a probability of 1.5% per antenna being defective, quality assurance teams can plan for expected levels of defects. This helps in:
- Predicting the number of defective units over a large batch.
- Scheduling inspections and implementing quality improvements.
- Pricing strategies, considering some goods might be unsellable or returned.
Exploring Random Sampling
Random sampling involves selecting a group in such a way that every individual item has an equal chance of being chosen. This method is crucial for ensuring data collected is representative of the larger population.
In the exercise, a random sample of 200 antennas is used to estimate the probability of defects. This sampling method ensures that each of the 200 antennas was equally likely to be chosen, providing a reliable basis for probability calculations.
Key advantages include:
In the exercise, a random sample of 200 antennas is used to estimate the probability of defects. This sampling method ensures that each of the 200 antennas was equally likely to be chosen, providing a reliable basis for probability calculations.
Key advantages include:
- Avoiding bias in sample selection, which means results are more credible.
- Enabling statistical inferences about the wider population from which the sample was drawn.
- Facilitating the use of probability models like the binomial distribution confidently.
