Normal Probability Distribution Using R

 

Experiment

Normal Probability Distribution Using R

1. Aim

To study and implement the Normal probability distribution in R by calculating probabilities, standard scores, generating probability values, and visualizing the normal distribution.

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2. Objectives

After completing this experiment, students should be able to:

1. Understand the concept and characteristics of the Normal distribution.
2. Identify the parameters of a Normal distribution.
3. Calculate probabilities using the Normal distribution formula.
4. Calculate the standard score ($z$-score).
5. Calculate probabilities using R.
6. Find probabilities below, above, and between specified values.
7. Generate values from a Normal distribution.
8. Visualize the Normal distribution using R.
9. Interpret the results obtained from a Normal distribution.

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3. Theory

3.1 Normal Distribution

The Normal distribution is a continuous probability distribution that is widely used to model naturally occurring measurements such as:

• Heights
• Weights
• Measurement errors
• Examination scores
• Manufacturing measurements
• Blood pressure
• Test scores

The probability density function of a Normal distribution is:

$z = \frac{x - \mu}{\sigma}$

$z=\frac{(\text{4})-(\text{1.5})}{\text{1.8}}=\text{1.389}$

$\Phi(z)\approx \text{91.8\%}$

$x$

$\mathrm{<br>}$

$\mu$

$\mathrm{<br>}$

$\sigma$

$\mathrm{<br>}$

$$f(x)=\frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$$

where:

• $\mu$ = population mean
• $\sigma$ = population standard deviation
• $x$ = observed value

A Normal distribution is completely determined by its mean $\mu$ and standard deviation $\sigma$.

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3.2 Characteristics of Normal Distribution

A Normal distribution has the following important characteristics:

1. It is continuous.
2. It is bell-shaped.
3. It is symmetric about the mean.
4. Mean, median, and mode are equal.
5. The total area under the curve is 1.
6. The curve extends theoretically from $-\infty$ to $+\infty$.
7. The standard deviation determines the spread of the distribution.

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4. Problem Statement

The marks obtained by students in a particular examination are approximately normally distributed with:

$$\mu=70$$

and

$$\sigma=10$$

Let $X$ represent the examination mark of a randomly selected student.

Therefore:

$$X\sim N(70,10^2)$$

Using the Normal distribution:

1. Calculate the $z$-score for a student who scored 80 marks.
2. Find the probability that a randomly selected student scores less than 80.
3. Find the probability that a student scores more than 80.
4. Find the probability that a student scores between 60 and 80.
5. Find the probability that a student scores between 70 and 90.
6. Generate a set of random examination marks from the Normal distribution.
7. Plot the Normal probability density curve.
8. Interpret the results.

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5. Manual Calculation

5.1 Calculation of the Z-score

The standard score or $z$-score is calculated using:

$$z=\frac{x-\mu}{\sigma}$$

For $x=80$:

$$z=\frac{80-70}{10}$$ $$\boxed{z=1}$$

Therefore, a score of 80 is one standard deviation above the mean.

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5.2 Probability of Scoring Less Than 80

We need to find:

$$P(X<80)$$

First convert 80 into a $z$-score:

$$z=\frac{80-70}{10}=1$$

Therefore:

$$P(X<80)=P(Z<1)$$

From the standard Normal distribution table:

$$P(Z<1)\approx0.8413$$

Therefore:

$$\boxed{P(X<80)\approx0.8413}$$

or approximately 84.13%.

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5.3 Probability of Scoring More Than 80

We need:

$$P(X>80)$$

Using the complement:

$$P(X>80)=1-P(X\leq80)$$

Therefore:

$$P(X>80)=1-0.8413$$ $$\boxed{P(X>80)\approx0.1587}$$

or approximately 15.87%.

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5.4 Probability of Scoring Between 60 and 80

We need:

$$P(60<X<80)$$

Calculate the two $z$-scores.

For $X=60$:

$$z_1=\frac{60-70}{10}=-1$$

For $X=80$:

$$z_2=\frac{80-70}{10}=1$$

Therefore:

$$P(60<X<80)=P(-1<Z<1)$$

From the standard Normal distribution:

$$P(Z<1)=0.8413$$

and

$$P(Z<-1)=0.1587$$

Therefore:

$$P(-1<Z<1)=0.8413-0.1587$$ $$\boxed{P(60<X<80)=0.6826}$$

or approximately 68.26%.

This illustrates the well-known 68% rule: approximately 68% of observations in a Normal distribution lie within one standard deviation of the mean.

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5.5 Probability of Scoring Between 70 and 90

We need:

$$P(70<X<90)$$

For 70:

$$z_1=\frac{70-70}{10}=0$$

For 90:

$$z_2=\frac{90-70}{10}=2$$

Therefore:

$$P(70<X<90)=P(0<Z<2)$$

Using the standard Normal distribution:

$$P(Z<2)\approx0.9772$$

and

$$P(Z<0)=0.5$$

Therefore:

$$P(0<Z<2)=0.9772-0.5$$ $$\boxed{P(70<X<90)\approx0.4772}$$

or approximately 47.72%.

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6. R Functions for Normal Distribution

R provides the following important functions:

R Function	Purpose
dnorm()	Probability density at a specified value
pnorm()	Cumulative probability
qnorm()	Quantile corresponding to a probability
rnorm()	Generates random observations

For a Normal distribution with mean 70 and standard deviation 10:

dnorm(x, mean = 70, sd = 10)

calculates the density at $x$.

pnorm(x, mean = 70, sd = 10)

calculates:

$$P(X\leq x)$$

qnorm(p, mean = 70, sd = 10)

finds the value $x$ corresponding to a specified cumulative probability.

rnorm(n, mean = 70, sd = 10)

generates n random observations from the Normal distribution.

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7. Program

# Normal Probability Distribution

# Parameters
mu <- 70
sigma <- 10


# 1. Calculate Z-score for X = 80

x <- 80

z <- (x - mu) / sigma

cat("Z-score for X = 80 =", z, "\n")


# 2. Probability of scoring less than 80

p_less_80 <- pnorm(80,
                   mean = mu,
                   sd = sigma)

cat("P(X < 80) =", p_less_80, "\n")


# 3. Probability of scoring more than 80

p_more_80 <- 1 - pnorm(80,
                        mean = mu,
                        sd = sigma)

cat("P(X > 80) =", p_more_80, "\n")


# 4. Probability of scoring between 60 and 80

p_60_80 <- pnorm(80,
                 mean = mu,
                 sd = sigma) -
           pnorm(60,
                 mean = mu,
                 sd = sigma)

cat("P(60 < X < 80) =", p_60_80, "\n")


# 5. Probability of scoring between 70 and 90

p_70_90 <- pnorm(90,
                 mean = mu,
                 sd = sigma) -
           pnorm(70,
                 mean = mu,
                 sd = sigma)

cat("P(70 < X < 90) =", p_70_90, "\n")


# 6. Generate random examination marks

set.seed(123)

marks <- rnorm(100,
               mean = mu,
               sd = sigma)

cat("\nFirst 10 generated marks:\n")
print(marks[1:10])


# 7. Plot the Normal distribution

x_values <- seq(30, 110, by = 0.5)

y_values <- dnorm(x_values,
                  mean = mu,
                  sd = sigma)

plot(x_values,
     y_values,
     type = "l",
     main = "Normal Probability Distribution",
     xlab = "Examination Marks",
     ylab = "Probability Density")

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8. Expected Result

The program produces approximately the following results:

Z-score for X = 80 = 1

P(X < 80) = 0.8413447

P(X > 80) = 0.1586553

P(60 < X < 80) = 0.6826895

P(70 < X < 90) = 0.4772499

The program also generates 100 random examination marks following:

$$N(70,10^2)$$

and displays the Normal distribution as a bell-shaped curve.

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9. Interpretation

From the results:

• A score of 80 has a $z$-score of 1, meaning it is one standard deviation above the mean.
• Approximately 84.13% of students are expected to score below 80.
• Approximately 15.87% are expected to score above 80.
• Approximately 68.27% are expected to score between 60 and 80.
• Approximately 47.72% are expected to score between 70 and 90.

The graph should show the characteristic symmetric bell-shaped curve of the Normal distribution, centered at the mean of 70.

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10. Result

Thus, the Normal probability distribution was studied and implemented using R. The z-score and probabilities for different ranges of examination marks were calculated manually using the standard Normal distribution concept and verified using R functions. Random observations were generated from the Normal distribution, and the corresponding probability density curve was plotted and interpreted.