/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 107 Forty percent of seeds from maiz... [FREE SOLUTION] | 91Ó°ÊÓ

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Forty percent of seeds from maize (modern-day corn) ears carry single spikelets, and the other \(60 \%\) carry paired spikelets. A seed with single spikelets will produce an ear with single spikelets \(29 \%\) of the time, whereas a seed with paired spikelets will produce an ear with single spikelets \(26 \%\) of the time. Consider randomly selecting ten seeds. a. What is the probability that exactly five of these seeds carry a single spikelet and produce an ear with a single spikelet? b. What is the probability that exactly five of the ears produced by these seeds have single spikelets? What is the probability that at most five ears have single spikelets?

Short Answer

Expert verified
a. 0.0106\n b. Exactly 5: 0.2301, At most 5: 0.8384.

Step by step solution

01

Determine Probability for Single Spikelets

We need the probability that a seed carries a single spikelet and produces an ear with single spikelets. A seed carries single spikelets with probability 0.4 and produces an ear with single spikelets with probability 0.29 given it carries single spikelets. Therefore, the overall probability is \[ P(A) = 0.4 \times 0.29 = 0.116 \]
02

Determine Probability for Paired Spikelets

Similarly, when a seed carries paired spikelets, the probability of producing an ear with single spikelets is 0.26.\[ P(B) = 0.6 \times 0.26 = 0.156 \]
03

Calculate Total Probability of Single Spikelet Ear

The probability that an ear has single spikelets can be obtained by summing the probabilities calculated in steps 1 and 2:\[ P(S) = P(A) + P(B) = 0.116 + 0.156 = 0.272 \]
04

Calculate Probability for Exactly Five Single Spikelets (a)

Use the binomial probability formula since we deal with a fixed number of trials (10 seeds) and two possible outcomes (having exactly 5 single spikelet results). The formula is:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \( n = 10 \), \( k=5 \), \( p = 0.116 \). So,\[ P(X = 5) = \binom{10}{5} (0.116)^5 (0.884)^5 \approx 0.0106 \]
05

Calculate Probability for Exactly Five Single Spikelet Ears (b)

Use the same binomial formula with \( p = 0.272 \) due to it being the total probability of an ear having single spikelets:\[ P(X = 5) = \binom{10}{5} (0.272)^5 (0.728)^5 \approx 0.2301 \]
06

Calculate Probability for At Most Five Single Spikelet Ears

Calculate the probability for 0 up to 5 single spikelet ears using \( p = 0.272 \) and summing those probabilities:\[ P(X \leq 5) = \sum_{k=0}^{5} \binom{10}{k} (0.272)^k (0.728)^{10-k} \approx 0.8384 \]

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Key Concepts

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

Binomial Distribution
The Binomial Distribution is a common statistical method used to model the probability of a fixed number of successes in a series of independent trials. These trials must have only two possible outcomes, commonly referred to as "success" or "failure."

In the context of our exercise, the success is defined as producing an ear with a single spikelet. The binomial distribution is characterized by two parameters, namely the number of trials (\( n \)) and the probability of success (\( p \)). Here, we have 10 trials, represented by the 10 seeds, and the probability is determined by earlier calculations.

This probability model offers crucial insights, as it allows us to make predictions about the occurrence of successful outcomes under given conditions. It forms the basis for determining probabilities such as exactly five seeds producing the desired outcome, as seen in the problem.
Probability Calculation
Probability Calculation forms the foundation of predicting outcomes in a binomial distribution setup. It involves computing probabilities using specific formulas that take account of known probabilities and desired outcomes.

The first step in our problem was calculating the probability of seeds with single and paired spikelets producing single spikelet ears. This is done by multiplying the probability of a seed type with the conditional probability of producing a single spikelet ear from that seed type.

For single spikelets, we calculated it as:\[ P(A) = 0.4 \times 0.29 = 0.116 \]For paired spikelets, it was:\[ P(B) = 0.6 \times 0.26 = 0.156 \]The combined probability of any seed producing a single spikelet ear was:\[ P(S) = P(A) + P(B) = 0.272 \]These probabilities set the stage for further binomial probability calculations, enabling us to address complex probability inquiries without relying only on intuition or guesswork.
Statistical Inference
Statistical Inference is the process used to make predictions or generalizations about a population based on a sample. In the context of our exercise, this principle is applied to infer the likelihood of obtaining certain outcomes when using seeds with various spikelet types.

With the derived probability \( P(S) \) for a seed producing a single spikelet ear, statistical inference comes into play when we evaluate the probabilities of different potential outcomes, such as exactly five or at most five single spikelet ears in 10 trials.

Leveraging the cumulative binomial probability calculation, we sum probabilities from 0 to 5 successes, which provides a more comprehensive understanding of likely outcomes. This approach of using statistical inference deeply enhances our prediction accuracy and decision-making, contributing significantly to fields where predictions based on sample data are vital. \[ P(X \leq 5) = \sum_{k=0}^{5} \binom{10}{k} (0.272)^k (0.728)^{10-k} \approx 0.8384 \]

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Most popular questions from this chapter

A manufacturer of flashlight batteries wishes to control the quality of its product by rejecting any lot in which the proportion of batteries having unacceptable voltage appears to be too high. To this end, out of each large lot ( 10,000 batteries), 25 will be selected and tested. If at least 5 of these generate an unacceptable voltage, the entire lot will be rejected. What is the probability that a lot will be rejected if a. \(5 \%\) of the batteries in the lot have unacceptable voltages? b. \(10 \%\) of the batteries in the lot have unacceptable voltages? c. \(20 \%\) of the batteries in the lot have unacceptable voltages? d. What would happen to the probabilities in parts (a) - (c) if the critical rejection number were increased from 5 to 6 ?

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