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Chapter 6: Problem 32

A pharmaceutical company knows that approximately \(5 \%\) of its birth-control pills have an ingredient that is below the minimum strength, thus rendering the pill ineffective. What is the probability that fewer than 10 in a sample of 200 pills will be ineffective?

Short Answer

Expert verified
The probability that fewer than 10 in a sample of 200 pills will be ineffective can be obtained by computing the cumulative binomial probability for 'p' = 0.05, 'n' = 200, and 'k' = 9. The exact numeric probability depends on the specific calculator or software used and should be performed by the student.

Step by step solution

01

Identify the Parameters of the Binomial Distribution

In this problem, the number of trials 'n' is the number of pills, which is 200. The probability of success 'p' is the probability that a pill is ineffective, which is \(0.05\). We want to find the probability 'P(k < 10)' of observing fewer than 10 successes.
02

Calculate the Binomial Probability

The formula for binomial probability is given by \( (n choose k) * (p^k) * ((1-p)^(n-k)) \). We need to employ this formula for k from 0 to 9 and sum up the results. This computation might be lengthy by hand but can be easily carried out with a suitable computer program, a scientific calculator, or a statistical software.
03

Find the cumulative probability

Most calculators and softwares have an in-built function for finding the cumulative binomial probability up to a certain number of successes 'k'. In this case, we want to find \(P(k < 10)\) which is same as \(P(k \leq 9)\). Using the built-in function for cumulative binomial probability, we can directly compute this value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability of Success
In a Binomial Distribution, the 'Probability of Success' refers to the chance that a single trial in a series of trials will result in a desired outcome. For the given problem, the probability of success is the likelihood that a birth-control pill is ineffective. Here, it is specified as 5%, or mathematically, \( p = 0.05 \). This is crucial because it sets the basis for calculating probabilities in multiple trials.
  • Probability of success \( (p) \) should always be a value between 0 and 1.
  • The probability of failure \( (1-p) \) is simply the complement of the probability of success.
Understanding this concept helps us model real-world situations using the Binomial Distribution, where outcomes are typically classified as 'success' (occurrence of the event) or 'failure' (non-occurrence of the event).
Cumulative Probability
Cumulative Probability is the sum of the probabilities of a range of outcomes in a distribution. In the context of the exercise, it's the probability of having fewer than 10 ineffective pills out of 200. This means summing the probabilities from 0 to 9 ineffective pills.
  • This involves calculating each individual probability using the binomial formula and then adding them together until you reach the desired number, which in this case is 9.
  • Mathematical tools and statistical software can automate this, providing cumulative probability functions to simplify calculations.
In practical terms, cumulative probabilities allow us to assess risks or chances by looking at summed outcomes rather than isolated instances. Here, it helps determine the overall chance of having low numbers of ineffective pills in a sample, rather than calculating each possibility individually.
Statistical Software
Using Statistical Software can greatly simplify calculations in statistics, especially when dealing with complex distributions like the Binomial Distribution. Tools such as R, Python's SciPy library, or even specialized calculators offer functions to compute the cumulative probability directly.
  • These softwares usually have built-in functions that calculate binomial probability and cumulative probability quickly.
  • They are very accurate and save time, especially compared to manual calculations.
  • To use them, one typically inputs the parameters: number of trials, probability of success, and the desired number of successes. The software then outputs the probability.
For students and professionals alike, familiarizing oneself with statistical software not only aids in homework but also provides valuable skills for data analysis in research and professional applications.

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Most popular questions from this chapter

Given a continuous uniform distribution, show that (a) \(\mu=\frac{A+B}{2},\) and (b) \(\sigma^{2}-\frac{(B-\lambda)^{2}}{12}\)

A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 milliliters, (a) what fraction of the cups will contain more than 224 milliliters? (b) what is the probability that a cup contains between 191 and 209 milliliters? (c) how many cups will probably overflow if 230 milliliter cups are used for the next 1000 drinks? (d) below what value do we get the smallest \(25 \%\) of the drinks?

Given a standard normal distribution. find the normal curve area under the curve which lies (a) to the left of \(z=1.43\); (b) to the right of \(z=-0.89\) : (c) between \(z=-2.16\) and \(z=-0.65\) (el) to the left of; \(=-1.39\); (e) to the: right of \(z=1.96\) : (T) between \(z=-0.48\) and \(z=1.74\).

The IQs of 600 applicants of a certain college are approximately normally distributed with a mean of 115 and a standard deviation of \(12 .\) If the college requires an IQ of at least, \(95,\) how many of these students will be rejected on this basis regardless of their other qualifications?

A manufacturer of a certain type of large machine wishes to buy rivets from one of two manufacturers. It is important that the breaking strength of each rivet exceed 10,000 psi. Two manufacturers (A and \(B\) ) offer this type of rivet and both have rivets whose breaking strength is normally distributed. The mean breaking strengths for manufacturers \(A\) and \(B\) are 14,000 psi and 13,000 psi, respectively. The standard deviations are 2000 psi and 1000 psi, respectively. Which manufacturer will produce, on the average, the fewest number of defective rivets?
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