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\(X\) is a normally distributed random variable with mean 112 and standard deviation \(15 .\) Find the probability indicated. a. \(\quad P(100

Short Answer

Expert verified
a. 0.5959, b. 0.2899, c. 0.3439

Step by step solution

01

Understand the Distribution

The random variable \(X\) follows a normal distribution with mean \(\mu = 112\) and standard deviation \(\sigma = 15\). This means we can use the properties of the normal distribution to solve for probabilities using the standard normal distribution (Z).
02

Find Z-scores for Problem a

For part (a), we want to find \(P(100 < X < 125)\). First, convert 100 and 125 into Z-scores:\[ Z = \frac{X - \mu}{\sigma} \]For \(X = 100\):\[ Z = \frac{100 - 112}{15} = -0.8 \]For \(X = 125\):\[ Z = \frac{125 - 112}{15} = 0.867 \]
03

Calculate Probability for Problem a

Using the Z-table, find the probabilities:\(P(Z < -0.8) \approx 0.2119\) and \(P(Z < 0.867) \approx 0.8078\).Thus, \(P(-0.8 < Z < 0.867) = P(Z < 0.867) - P(Z < -0.8) = 0.8078 - 0.2119 = 0.5959\).
04

Find Z-scores for Problem b

For part (b), we need \(P(91 < X < 107)\). Convert 91 and 107 into Z-scores:For \(X = 91\):\[ Z = \frac{91 - 112}{15} = -1.4 \]For \(X = 107\):\[ Z = \frac{107 - 112}{15} = -0.3333 \]
05

Calculate Probability for Problem b

Using the Z-table, find the probabilities:\(P(Z < -1.4) \approx 0.0808\) and \(P(Z < -0.3333) \approx 0.3707\).Thus, \(P(-1.4 < Z < -0.3333) = P(Z < -0.3333) - P(Z < -1.4) = 0.3707 - 0.0808 = 0.2899\).
06

Find Z-scores for Problem c

For part (c), we need \(P(118 < X < 160)\). Convert 118 and 160 into Z-scores:For \(X = 118\):\[ Z = \frac{118 - 112}{15} = 0.4 \]For \(X = 160\):\[ Z = \frac{160 - 112}{15} = 3.2 \]
07

Calculate Probability for Problem c

Using the Z-table, find the probabilities:\(P(Z < 0.4) \approx 0.6554\) and \(P(Z < 3.2) \approx 0.9993\).Thus, \(P(0.4 < Z < 3.2) = P(Z < 3.2) - P(Z < 0.4) = 0.9993 - 0.6554 = 0.3439\).

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Key Concepts

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

Probability Calculation
When working with normal distributions, calculating probabilities involves finding the area under the curve for a specific range of values.
The goal is to determine how likely a variable will fall between two points. For the given problem involving the random variable \(X\), we need to find the probabilities of \(X\) falling between certain values by converting these values to corresponding Z-scores.
Here's how we calculate the probability using the standard normal distribution (Z-distribution):
  • Identify the specific range you are interested in, such as \(100 < X < 125\).
  • Convert the boundary values to Z-scores to standardize them.
  • Use the Z-table to find the area under the curve for each Z-score, which represents the probability of values being less than that Z-score.
  • Subtract the smaller probability from the larger one to find the probability of being within the interval.
These steps allow you to determine the probability, ensuring a clear understanding of how likely it is for a value to land in a particular range.
Z-scores
Z-scores are critical in normal distribution problems because they allow you to standardize any normal distribution to a standard normal distribution, which has a mean of 0 and a standard deviation of 1. This makes it easier to use pre-existing Z-tables to find probabilities.
Here's how to convert a raw score to a Z-score:
  • Subtract the mean \((\mu)\) of the distribution from the raw score \(X\).
  • Divide the result by the standard deviation \((\sigma)\).
Mathematically, the formula is given by : \[ Z = \frac{X - \mu}{\sigma} \]This conversion simplifies computations and allows comparison across different normal distributions.
In our problem, converting \(X = 100\) and \(X = 125\) into Z-scores helps in finding the standard probabilities since these scores directly link to specific probabilities in the Z-table.
Standard Normal Distribution
The standard normal distribution is a special type of normal distribution where the mean \, \((\mu)\, = 0\) and the standard deviation \((\sigma)\, = 1\).
Points on this bell-shaped curve can be described using Z-scores that tell us how far a point is from the mean in terms of standard deviations.
Why is it important?
  • The standard normal distribution is universally recognized and provides a universal standard for finding probabilities.
  • Using this distribution ensures consistency and allows us to leverage Z-tables, which specifically give cumulative probability outcomes for various Z-scores.
By using Z-tables or software tools, we can quickly determine the probability of a given Z-score, making complex calculations manageable.
In exercises like these, assumptions are made about using this standardized form to solve otherwise intricate probability problems.

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Most popular questions from this chapter

A machine for filling 2 -liter bottles of soft drink delivers an amount to each bottle that varies from bottle to bottle according to a normal distribution with standard deviation 0.002 liter and mean whatever amount the machine is set to deliver. a. If the machine is set to deliver 2 liters (so the mean amount delivered is 2 liters) what proportion of the bottles will contain at least 2 liters of soft drink? b. Find the minimum setting of the mean amount delivered by the machine so that at least \(99 \%\) of all bottles will contain at least 2 liters.

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