# Gaussian Distribution

sorry to bother people with this, but my stats teacher did not make solving gaussian distribution questions, clear AT ALL in his notes, and my exam is coming soon, someone explain to me what is happening in here (like how the final answer is found)

P (X < 95.5) = P( (X − 98.8)/2 <( 95.5 − 98.8)/ 2) = P (Z < −1.65) = 0.0495

(heres the full question where the above was the solution)

1. A machine is used to fill tubes, of nominal content 100 ml, with toothpaste. The amount of toothpaste delivered by the machine is normally distributed and may be set to any required mean value. Immediately after the machine has been over- hauled, the standard deviation of the amount delivered is 2 ml. As time passes, this standard deviation increases until the machine is again overhauled. The following three conditions are necessary for a batch of tubes of toothpaste to comply with current legislation: • I: the average content of the tubes must be at least 100 ml, • II: not more than 2.5% of the tubes may contain less than 95.5 ml, • III: not more than 0.1% of the tubes may contain less than 91 ml.

(a) For a batch of tubes with mean content 98.8 ml and standard deviation 2 ml,

find the proportion of tubes which contain i. less than 95.5 ml, Solution: Let X = amount of toothpaste in a tube (in ml). Then X ∼ G(98.8, 2).

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This looks like a problem that says something like

"The body temperature of a human being can be modeled as a Gaussian random variable $X$ with mean $\mu=98.8$ and standard deviation $\sigma=2$. What is the probability that the body temperature is less than $95.5$?"

What you need for your exam is to know that $P\{X<a\}$ is found by computing $z=(a-\mu)/\sigma$ and then looking in a table of values of $\Phi(\cdot)$, the cumulative standard normal distribution, for the value of $\Phi(z)$ (or plugging in the numbers into your calculator if calculators are allowed on the exam). Tables of $\Phi(\cdot)$ often list the values for positive arguments only, and so you also need to know that $\Phi(-z) = 1 - \Phi(z)$, and so $\Phi(-1.65) = 1-\Phi(1.65)$.

Edit The actual problem has now been included in the question. It wasn't when I wrote my answer.

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THANK YOU!!!!!!! – math_guy Feb 17 '12 at 14:20

Here is a recipe:

Let $\Phi$ denote the distribution function of $A \sim N(0,1)$. We'd like to know the distribution function of the random variable, $X \sim N(\mu, \sigma^2)$. Let $x \in \mathbb R$

\begin{align}P(X \le x)&=P(\mu + \sigma A \le x)\\&=P(A \le \dfrac{x -\mu}{\sigma})\\&=\Phi(\dfrac{x-\mu}{\sigma})\end{align}

Let's fix arbitrary real numbers $a, b \in \mathbb R$ such that $a \le b$. From the previous calculation,

$$P(a \le X \le b)=\Phi(\dfrac{b-\mu}{\sigma})-\Phi(\dfrac{a-\mu}{\sigma})$$

What is $A$ here is referred to as $Z$ in standard notation. And, $\Phi(z)$ is tabulated in various forms in the Data books.

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