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Chapter 15: Problem 36

A company selling vegetable seeds in packets of 20 estimates that the mean number of seeds that will actually grow is \(18,\) with a standard deviation of 1.2 seeds. You buy 5 different seed packets. a) How many good seeds do you expect to get? b) What's the standard deviation? c) What assumptions did you make about the seeds? Do you think that assumption is warranted? Explain.

Short Answer

Expert verified
a) 90 seeds expected. b) Standard deviation is approximately 2.68. c) Assumes independent distribution; reasonable if seeds are consistently produced.

Step by step solution

01

Calculate the Expected Total Number of Good Seeds

First, recognize that each packet has an expected number of 18 good seeds. If 5 packets are purchased, the expected number of good seeds is the sum of expected seeds in each packet. Mathematically, this is given by multiplying the expected value per packet by the number of packets: \(18 \times 5 = 90\). So, you expect to get 90 good seeds in total.
02

Find the Standard Deviation for Total Number of Good Seeds

Next, recognize that when dealing with independent random variables (as with each packet), the standard deviation of the total is found using the formula for the sum of variances. For each packet, the standard deviation is 1.2, so the variance is \(1.2^2 = 1.44\). For 5 packets, the total variance is \(5 \times 1.44 = 7.2\). The standard deviation for all 5 packets is the square root of the total variance: \(\sqrt{7.2} \approx 2.68\).
03

Discuss the Assumptions Made

The assumption made here is that the number of good seeds follows a normal distribution or some similar distribution across packets, which means independence between the packets. This implies that the number of seeds performing well in one packet does not affect the others. This assumption might be reasonable depending on the production process consistency and environmental conditions. It assumes that each packet鈥檚 growing conditions and factors are independent of the other packets.

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Key Concepts

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

Expected Value
In probability and statistics, the expected value (or mean) is a crucial concept. It tells us the average outcome we expect in a random experiment. It's like forecasting the most likely result of a situation if we could repeat it countless times.

In the exercise given, each seed packet has an expected number of 18 good seeds. Since you have 5 packets, you simply multiply the expected number by the total packets:
  • Expected good seeds per packet = 18
  • Total packets = 5
  • Expected total good seeds = 18 脳 5 = 90
This calculation uses a simple multiplication principle of expected values, which is quite handy with independent events like these.
Standard Deviation
Standard deviation measures how spread out the numbers in a data set are. It tells us how much we might expect something to deviate from the average, giving insight into variability.

From the exercise, we know that the standard deviation for one packet is 1.2 seeds. To find the combined standard deviation across multiple independent packets, we must first find the total variance. Variance is the square of the standard deviation:
  • Variance per packet = \(1.2^2 = 1.44\)
  • Total variance for 5 packets = 5 脳 1.44 = 7.2
The square root of the total variance gives the total standard deviation:
  • Combined standard deviation = \sqrt{7.2} \(\approx 2.68\)
This helps gauge the expected fluctuation in seed growth across multiple packets.
Independence Assumption
Independence is a key assumption when calculating expected values and standard deviations for a group of random variables. In our context, it means the number of seeds that grow in one packet does not affect another packet.

When packets are independent, each packet is treated individually without influencing the others. This assumption allows us to directly sum their expected values and variances.
  • Independence implies no crossover of conditions between packets.
  • Each packet behaves like a separate experiment.
Assuming independence might be reasonable if the production and storage of each packet are consistent, and environmental variables such as soil and watering don't vary much across packets.
Random Variables
Random variables are a foundational concept in probability, representing outcomes of random phenomena. Each packet, in our example, represents a random variable reflecting the successful seed growth.

Every random variable has its probability distribution representing the likelihood of different outcomes. For the seeds,
  • Each packet contains a number of good seeds, a random variable with a mean (expected value) and standard deviation.
Random variables allow us to apply mathematical principles to real-world uncertainties. By understanding each packet as a random variable, we describe the likely results of growing seeds based on statistical properties.

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Most popular questions from this chapter

Given independent random variables with means and standard deviations as shown, find the mean and standard deviation of: $$\begin{array}{c|c|c} & \text { Mean } & \text { SD } \\\\\hline x & 80 & 12 \\\Y & 12 & 3\end{array}$$ a) \(X-20\) b) \(0.5 Y\) c) \(X+Y\) d) \(X-Y\) e) \(Y_{1}+Y_{2}\)

An insurance company estimates that it should make an annual profit of \(\$ 150\) on each homeowner's policy written, with a standard deviation of \(\$ 6000\) a) Why is the standard deviation so large? b) If it writes only two of these policies, what are the mean and standard deviation of the annual profit? c) If it writes 10,000 of these policies, what are the mean and standard deviation of the annual profit? d) Is the company likely to be profitable? Explain. e) What assumptions underlie your analysis? Can you think of circumstances under which those assumptions might be violated? Explain.

At a certain coffee shop, all the customers buy a cup of coffee; some also buy a doughnut. The shop owner believes that the number of cups he sells each day is normally distributed with a mean of 320 cups and a standard deviation of 20 cups. He also believes that the number of doughnuts he sells each day is independent of the coffee sales and is normally distributed with a mean of 150 doughnuts and a standard deviation of 12. a) The shop is open every day but Sunday. Assuming day-to-day sales are independent, what's the probability he'll sell over 2000 cups of coffee in a week? b) If he makes a profit of 50 cents on each cup of coffee and 40 cents on each doughnut, can he reasonably expect to have a day's profit of over \(\$ 300 ?\) Explain. c) What's the probability that on any given day he'll sell a doughnut to more than half of his coffee customers?

In a litter of seven kittens, three are female. You pick two kittens at random. a) Create a probability model for the number of male kittens you get. b) What's the expected number of males? c) What's the standard deviation?

A man buys a racehorse for \(\$ 20,000\) and enters it in two races. He plans to sell the horse afterward, hoping to make a profit. If the horse wins both races, its value will jump to \(\$ 100,000 .\) If it wins one of the races, it will be worth \(\$ 50,000 .\) If it loses both races, it will be worth only \(\$ 10,000 .\) The man believes there's a \(20 \%\) chance that the horse will win the first race and a \(30 \%\) chance it will win the second one. Assuming that the two races are independent events, find the man's expected profit.
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