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Car colour preferences change over time and according to the area the customer lives in and the car model he/she is interested in. In a certain city, a large dealer of BMW cars noticed that \(10 \%\) of the cars he sells are 'metallic grey? Twenty of his customers are selected at random, and their car orders are checked for colour. Find the probability that a) at least five cars are'metallic grey' b) at most 6 cars are 'metallic grey" c) more than 5 are 'metallic grey' d) between 4 and 6 are 'metallic grey' e) more than 15 are not'metallic grey: In a sample of 100 customer records, find f) the expected number of'metallic grey'car orders g) the standard deviation of'metallic grey'car orders. According to the empirical rule, \(95 \%\) of the 'metallic grey' orders are between \(a\) and \(b\) h) Find \(a\) and \(b\)

Short Answer

Expert verified
a) Use \(1 - P(X < 5)\) b) Use \(P(X \leq 6)\) c) Use \(1 - P(X \leq 5)\) d) Use \(P(4 \leq X \leq 6)\) e) Use \(P(X \leq 4)\) f) 10 g) Calculate \(\sigma\) h) Use \(10 \pm 2\sigma\)

Step by step solution

01

Define the Problem

We have a binomial distribution problem where the probability of success (choosing a metallic grey car) is \( p = 0.1 \) and the number of trials is \( n = 20 \) for cases (a) to (e) and \( n = 100 \) for case (f) to (h).
02

Probability for a) At least five

To find the probability of at least 5 metallic grey cars, calculate \( P(X \geq 5) \), where \( X \) follows a binomial distribution with parameters \( n = 20 \) and \( p = 0.1 \). Use the formula: \( P(X \geq 5) = 1 - P(X < 5) \), and sum probabilities \( P(X = 0) \) to \( P(X = 4) \).
03

Probability for b) At most six

To find the probability of at most 6 metallic grey cars, calculate \( P(X \leq 6) \), sum the probabilities from \( P(X = 0) \) to \( P(X = 6) \).
04

Probability for c) More than five

To find the probability of more than 5 metallic grey cars, calculate \( P(X > 5) = 1 - P(X \leq 5) \). First find \( P(X \leq 5) \) by summing probabilities up to \( P(X = 5) \).
05

Probability for d) Between 4 and 6

To find the probability of between 4 and 6 metallic grey cars, calculate \( P(4 \leq X \leq 6) \) by adding \( P(X = 4) + P(X = 5) + P(X = 6) \).
06

Probability for e) More than 15 not grey

For a sample of 20, calculate the probability of more than 15 cars not being metallic grey: equivalent to \( P(X \leq 4) \) for metallic grey, because \( 20 - X > 15 \) implies \( X < 5 \).
07

Expected Value for f)

To find the expected number of metallic grey car orders: \( E(X) = n \times p \). For \( n = 100 \) and \( p = 0.1 \), \( E(X) = 100 \times 0.1 = 10 \).
08

Standard Deviation for g)

The standard deviation \( \sigma \) is found using \( \sigma = \sqrt{n \times p \times (1-p)} \). For \( n = 100 \) and \( p = 0.1 \), \( \sigma = \sqrt{100 \times 0.1 \times 0.9} \).
09

Empirical Rule for h)

For the empirical rule, \( 95\% \) of values fall within \( \pm 2\sigma \) of the mean. So the range is \( [E(X) - 2\sigma, E(X) + 2\sigma] \). Calculate, using \( E(X) = 10 \) and \( \sigma \) from Step 8, to find \( a \) and \( b \).

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Key Concepts

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

Probability Calculations
Probability calculations are essential when dealing with binomial distributions. The probability helps determine the likelihood of various outcomes when conducting trials, such as flipping a coin or, in this context, selling a metallic grey BMW. In a binomial distribution, you have a fixed number of trials, each with two possible outcomes: success or failure. In this exercise, success means selling a metallic grey car with a probability of 0.1, and the number of trials is 20 or 100, depending on the scenario.

To find the probability for a particular range or number of successes, we can use the cumulative probability method. This involves summing individual probabilities of achieving 0 up to a certain number of successes:
  • At least five successes (\(P(X \geq 5)\)), subtract the sum of probabilities of having fewer than five successes from 1.
  • For at most six successes (\(P(X \leq 6)\)), add up probabilities from zero up to six successes.
  • More than five successes signified by subtracting probabilities for five or fewer (\(P(X > 5) = 1 - P(X \leq 5)\)).
  • Between 4 and 6 successes requires adding probabilities from four to six (\(P(4 \leq X \leq 6)\)).
  • More than 15 not being grey equals to fewer than 5 grey (inverse probability).
Knowing these calculations is crucial for interpreting random events with a binary outcome, as seen in the dealership context.
Expected Value
The expected value, denoted as \(E(X)\), is a fundamental concept in probability and statistics. Think of it as the 'average' you'd expect when you repeat an experiment many times. For a binomial distribution, the expected value is calculated using the formula: \(E(X) = n \times p\), where \(n\) is the number of trials, and \(p\) is the probability of success.

For example, if a BMW dealer sells metallic grey cars with a success probability of 0.1 over 100 trials, then:\[E(X) = 100 \times 0.1 = 10\] This means out of 100 cars sold, you expect 10 to be metallic grey on average. Understanding expected value gives you insight into the long-term outcome of random trials, which helps in decision-making and planning in real-world scenarios. Expected values are particularly helpful when assessing likely outcomes in fields like economics, finance, and quality assurance.
Standard Deviation
Standard deviation measures how spread out numbers are in a data set. In the context of binomial distribution, it tells you how much deviation to expect from the average number of successes. A larger standard deviation means more variability in the outcomes, while a smaller one indicates that most outcomes are close to the expected value.

To find the standard deviation in a binomial context, use the formula: \[\sigma = \sqrt{n \times p \times (1-p)}\] For the dealer example with \(n = 100\) and \(p = 0.1\), the standard deviation is calculated as follows: \[\sigma = \sqrt{100 \times 0.1 \times 0.9}\] Calculating this gives approximately 3. This means that the number of cars sold as metallic grey will typically vary by about 3 from the average of 10.

Understanding standard deviation helps in determining the consistency and predictability of your data. In our car dealership context, it helps manage stock and set realistic sales targets.
Empirical Rule
The Empirical Rule, often called the 68-95-99.7 rule, is a key concept in understanding data distributions, particularly when they are roughly normal. This rule states that for a normal distribution:
  • 68% of data falls within one standard deviation (\(\sigma\)) of the mean (\(\mu\)).
  • 95% falls within two \(\sigma\).
  • 99.7% falls within three \(\sigma\).
For our binomial problem with metallic grey cars, we apply the empirical rule to estimate that 95% of observations fall within two standard deviations from the mean.

Using the calculated expected value \(E(X)=10\) and standard deviation \(\sigma \approx 3\), the range for 95% of orders is: \[[E(X) - 2\sigma, E(X) + 2\sigma]\] This equates to: \[[10 - 2(3), 10 + 2(3)] = [4, 16]\] Thus, you can expect that about 95% of the time, the number of metallic grey car orders will fall between 4 and 16. This helps dealers in planning and setting realistic expectations for inventory and sales forecasting.

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Most popular questions from this chapter

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