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Approximately \(3.6 \%\) of all (untreated) Jonathan apples had bitter pit in a study conducted by the botanists Ratkowsky and Martin (Source: Australian Journal of Agricultural Research, Vol. 25, pp. \(783-790) .\) (Bitter pit is a disease of apples resulting in a soggy core, which can be caused either by overwatering the apple tree or by a calcium deficiency in the soil.) Let \(n\) be a random variable that represents the first Jonathan apple chosen at random that has bitter pit. (a) Write out a formula for the probability distribution of the random variable \(n .\) (b) Find the probabilities that \(n=3, n=5\), and \(n=12\). (c) Find the probability that \(n \geq 5\). (d) What is the expected number of apples that must be examined to find the first one with bitter pit? Hint: Use \(\mu\) for the geometric distribution and round.

Short Answer

Expert verified
(a) Use \( P(n) = (1-p)^{n-1}p \); (b) \( P(3)=0.0335 \), \( P(5)=0.0313 \), \( P(12)=0.0241 \); (c) \( P(n \geq 5)=0.8636 \); (d) Expected apples: 28.

Step by step solution

01

Identify the distribution

The problem states that we are looking for the number of trials needed to get the first success (an apple with bitter pit). This is a geometric distribution, which models the number of Bernoulli trials needed to get one success.
02

Geometric Distribution Formula

For a geometric distribution, the probability that the first success occurs on the nth trial is given by the formula \( P(n) = (1-p)^{n-1}p \), where \( p \) is the probability of success on each trial. Here, \( p = 0.036 \).
03

Calculate Specific Probabilities

Use the geometric distribution formula to calculate for \( n=3, n=5 \), and \( n=12 \).- For \( n=3 \): \( P(3) = (1-0.036)^{3-1} \times 0.036 = 0.0335 \).- For \( n=5 \): \( P(5) = (1-0.036)^{5-1} \times 0.036 = 0.0313 \).- For \( n=12 \): \( P(12) = (1-0.036)^{12-1} \times 0.036 = 0.0241 \).
04

Calculate Probability for n ≥ 5

To find the probability that \( n \geq 5 \), we sum the probabilities: \( P(n \geq 5) = 1 - P(n < 5) = 1 - (P(1) + P(2) + P(3) + P(4)) \).- Calculate each: \( P(1) = 0.036 \), \( P(2) = 0.0347 \), \( P(3) = 0.0335 \), \( P(4) = 0.0322 \).- Thus, \( P(n \geq 5) = 1 - (0.036 + 0.0347 + 0.0335 + 0.0322) = 0.8636 \).
05

Expected Number of Apples

The expected number for a geometric distribution is given by \( \mu = \frac{1}{p} \). With \( p = 0.036 \), the expected number of apples is \( \mu = \frac{1}{0.036} \approx 27.78 \). Round to the nearest whole number, yielding 28.

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Key Concepts

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

Probability distribution
In statistics, a probability distribution is a function that describes the likelihood of different outcomes in a random experiment. For specific values, the probability distribution assigns a probability to each possible outcome. In the case of discrete random variables, like those involving the count of trials until a success, probabilities are assigned to exact points (e.g., 3, 5, or 12 trials). This contrasts with continuous random variables, where probabilities are assigned to intervals. A probability distribution can be visualized as a graph where each outcome has a corresponding probability height. In this exercise, we specifically deal with a geometric distribution — a type of probability distribution suited to model scenarios that involve counting the trials until the first success is recorded. The geometric probability distribution is applicable when we're considering Bernoulli trials, which are experiments that result in a binary outcome: success or failure.
Bernoulli trials
Bernoulli trials are a sequence of random experiments where each experiment results in one of two outcomes: success or failure. These trials are characterized by the condition that each trial is independent of others and has the same probability of success.For example, in our exercise with Jonathan apples, treating an apple as a single trial, each examination of an apple can result in a 'success' (the apple has bitter pit) or a 'failure' (the apple does not have bitter pit).The probability of success is denoted by \( p \), and in this exercise, \( p = 0.036 \) or 3.6%. This means that for any given apple, there is a 3.6% chance it will have bitter pit. When Bernoulli trials are repeated until the first success, the number of trials needed follows a geometric distribution.
Expected value
The expected value is a key concept in probability, representing the long-term average or mean value of a random variable's outcomes over numerous trials. In the context of geometric distribution, the expected value is the average number of trials required to obtain the first success.For a geometric distribution, the expected value, denoted \( \mu \), is calculated as \( \mu = \frac{1}{p} \), where \( p \) is the probability of success. With \( p = 0.036 \), the expected value of trials needed to find the first Jonathan apple with bitter pit is approximately 28. This expected number reflects that, on average, one must examine around 28 apples to encounter one with bitter pit.This concept of expected value helps to understand and predict the 'typical' number of trials necessary, offering insights into how often one might experience a particular situation.
Bitter pit in apples
Bitter pit is a disorder affecting apples, where the fruit develops sunken, discolored spots and often has a soggy core. It primarily occurs due to calcium deficiency in the soil, which affects the apple's cell structure. Overwatering apple trees can exacerbate this condition. In the study referenced in our exercise, botanists determined that about 3.6% of untreated Jonathan apples exhibited signs of bitter pit. This disorder is significant in agricultural research and production because it adversely affects the quality and marketability of apples. Understanding the occurrence rate of bitter pit (3.6% in this case) allows farmers and researchers to predict how frequently they might encounter affected apples in a random sampling. This percentage forms the basis for calculating probabilities using the geometric distribution, as seen in the given exercise.

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Most popular questions from this chapter

Bob is a recent law school graduate who intends to take the state bar exam. According to the National Conference on Bar Examiners, about \(57 \%\) of all people who take the state bar exam pass (Source: The Book of \(\mathrm{Odds}\) by Shook and Shook, Signet). Let \(n=1,2,3, \ldots\) represent the number of times a person takes the bar exam until the first pass. (a) Write out a formula for the probability distribution of the random variable \(n\). (b) What is the probability that Bob first passes the bar exam on the second try \((n=2) ?\) (c) What is the probability that Bob needs three attempts to pass the bar exam? (d) What is the probability that Bob needs more than three attempts to pass the bar exam? (e) What is the expected number of attempts at the state bar exam Bob must make for his (first) pass? Hint: Use \(\mu\) for the geometric distribution and round.

Caribbean Cruise The college student senate is sponsoring a spring break Caribbean cruise raffle. The proceeds are to be donated to the Samaritan Center for the Homeless. A local travel agency donated the cruise, valued at \(\$ 2000 .\) The students sold 2852 raffle tickets at \(\$ 5\) per ticket. (a) Kevin bought six tickets. What is the probability that Kevin will win the spring break cruise to the Caribbean? What is the probability that Kevin will not win the cruise? (b) Expected earnings can be found by multiplying the value of the cruise by the probability that Kevin will win. What are Kevin's expected earnings? Is this more or less than the amount Kevin paid for the six tickets? How much did Kevin effectively contribute to the Samaritan Center for the Homeless?

Norb and Gary are entered in a local golf tournament. Both have played the local course many times. Their scores are random variables with the following means and standard deviations. $$ \text { Norb, } x_{1}: \mu_{1}=115 ; \sigma_{1}=12 \quad \text { Gary, } x_{2}: \mu_{2}=100 ; \sigma_{2}=8 $$ In the tournament, Norb and Gary are not playing together, and we will assume their scores vary independently of each other. (a) The difference between their scores is \(W=x_{1}-x_{2} .\) Compute the mean, variance, and standard deviation for the random variable \(W\). (b) The average of their scores is \(W=0.5 x_{1}+0.5 x_{2}\). Compute the mean, variance, and standard deviation for the random variable W (c) The tournament rules have a special handicap system for each player. For Norb, the handicap formula is \(L=0.8 x_{1}-2 .\) Compute the mean, variance, and standard deviation for the random variable \(L .\) (d) For Gary, the handicap formula is \(L=0.95 x_{2}-5 .\) Compute the mean, variance, and standard deviation for the random variable \(L\).

This problem will be referred to in the study of control charts (Section 6.1). In the binomial probability distribution, let the number of trials be \(n=3\), and let the probability of success be \(p=0.0228\). Use a calculator to compute (a) the probability of two successes. (b) the probability of three successes. (c) the probability of two or three successes.

For a binomial experiment, what probability distribution is used to find the probability that the first success will occur on a specified trial?

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