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91Ó°ÊÓ

A continuous random variable \(X\) has a normal distribution with mean 73 . The probability that \(X\) takes a value greater than 80 is \(0.212 .\) Use this information and the symmetry of the density function to find the probability that \(X\) takes a value less than \(66 .\) Sketch the density curve with relevant regions shaded to illustrate the computation.

Short Answer

Expert verified
P(X < 66) = 0.212, using the symmetry of the normal distribution.

Step by step solution

01

Understand the Problem

We are given a normal distribution with mean \( \mu = 73 \) and need to find the probability that \( X \) is less than 66. We know that \( P(X > 80) = 0.212 \).
02

Utilize Symmetry of the Normal Distribution

Normal distributions are symmetric about the mean. The properties of the normal distribution and its symmetry tell us that the probability of \( X \) being greater than a certain value above the mean is the same as being less than the same value below the mean. Thus, \( P(X < 66) = P(X > 80) \).
03

Calculate the Desired Probability

From Step 2, we have \( P(X < 66) = P(X > 80) = 0.212 \). So, the probability that \( X \) takes a value less than 66 is 0.212.
04

Sketch the Density Function

On a normal distribution bell curve with mean 73, mark the point at \( X = 73 \). Illustrate the region \( X > 80 \) on the right side of the curve, which is 0.212 shaded blue. Similarly, shade the region \( X < 66 \) on the left side of the curve with 0.212 in green. Both shaded areas represent the same probability due to symmetry.

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Key Concepts

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

Continuous Random Variable
A continuous random variable is a type of variable that can take an infinite number of values. This is different from discrete random variables, which can only take specific values.
In the context of the normal distribution, which is key here, a continuous random variable can assume any value along the entire range of real numbers.
Think about measuring height, weight, or temperature. These are all examples of continuous data because they can take any value within a range, and we can have as many decimal points as we need.
  • Continuous random variables are often modeled using distributions like the normal distribution.
  • They have associated probabilities for ranges of values, rather than specific values.
Ultimately, when we speak of a normal distribution in terms of continuous random variables, we are discussing the probability distribution over all possible real numbers, which is defined by certain parameters like the mean and standard deviation.
Mean
The mean, often denoted by the symbol \( \mu \), is a measure of central tendency. It represents the average value in a set of data.
In the context of a normal distribution, the mean is the peak or the center of the bell-shaped curve. Every continuous random variable with a normal distribution has a mean that is located at the exact midpoint of the curve, dividing it into two symmetrical halves.
  • In our specific problem, the mean is given as 73. This indicates that the average value expected for this distribution is 73.
  • The mean determines the location of the curve along the x-axis.
The mean not only helps in predicting the "average" scenario, but it also plays a vital role in the concept of symmetry, which we'll touch on next. Ultimately, knowing the mean gives us a solid reference point from which to understand the distribution's behavior.
Symmetry
Symmetry in a normal distribution plays a crucial role in simplifying probability calculations. The normal distribution is noted for its characteristic bell shape and perfect symmetry around the mean.
Imagine folding a piece of paper in half; if both sides match perfectly, it's symmetrical. The normal distribution works the same way around the mean.
  • For a mean of 73, the distribution will look the same on both sides.
  • This symmetry means that the probability of being a certain distance above the mean is equal to being the same distance below the mean.
In terms of our problem, knowing the symmetry allows us to equate the probability of \( X > 80 \) and \( X < 66 \), simplifying our calculations by using one known probability to find the other. This symmetrical property is why the probability of exceeding a certain threshold on one side of the mean can directly give us the probability on the opposite side.
Probability
Probability is a measure of how likely an event is to occur, often represented as a number between 0 and 1. A probability of 0 means the event will not occur, while a probability of 1 means the event will certainly occur.
In continuous distributions, like the normal distribution at hand, probability can be understood in terms of calculating the area under the curve.
  • When we say \( P(X > 80) = 0.212 \), it means that there's a 21.2% chance for our random variable \( X \) to take on a value greater than 80.
  • Similarly, the probability \( P(X < 66) = 0.212 \) reflects the chance of \( X \) being less than 66, based on the area under the curve to the left of this point.
Probabilities in a continuous distribution allow us to calculate the likelihood of a variable falling within certain intervals rather than exact values, due to its continuous nature. Using concepts like symmetry, we can determine probabilities without needing additional information, using known parts of the distribution.

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Most popular questions from this chapter

All students in a large enrollment multiple section course take common in- class exams and a common final, and submit common homework assignments. Course grades are assigned based on students' final overall scores, which are approximately normally distributed. The department assigns a C to students whose scores constitute the middle \(2 / 3\) of all scores. If scores this semester had mean 72.5 and standard deviation \(6.14,\) find the interval of scores that will be assigned a \(\mathrm{C}\)

A regulation hockey puck must weigh between 5.5 and 6 ounces. In an alternative manufacturing process the mean weight of pucks produced is 5.75 ounce. The weights of pucks have a normal distribution whose standard deviation can be decreased by increasingly stringent (and expensive) controls on the manufacturing process. Find the maximum allowable standard deviation so that at most 0.005 of all pucks will fail to meet the weight standard.

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Tests of a new tire developed by a tire manufacturer led to an estimated mean tread life of 67,350 miles and standard deviation of 1,120 miles. The manufacturer will advertise the lifetime of the tire (for example, a " 50,000 mile tire") using the largest value for which it is expected that \(98 \%\) of the tires will last at least that long. Assuming tire life is normally distributed, find that advertised value.

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