In this section, we discuss Normal Distribution, a continuous random variable distribution.

• Probability Distribution of a Continuous Random Variable
  
• The Characteristics of Normal Distribution (shape, mean, standard deviation)
  
• The Standard normal distribution
  
• Find probability for standard/general normal distribution using R
  
• Compute z-scores for any value of normally distributed random variable
  
• Find the value of normal distribution variable given probability
  
• Find percentile
  
• The 68-95-99.7 Rule (Empirical Rule)

- A random variable is called continuous when its possible values form an interval.
- The probability distribution of a continuous random variable \(X\) is denoted by a probability density function (pdf) \(𝑓(𝑥)\), with \(𝑓(𝑥)≥0\).
- The pdf of \(X\) is usually represented by a probability density curve over the interval of all possible values of the random variable as shown below.
- The probability that \(X\) falls between two values 𝑎 and 𝑏, is the area under the curve between 𝑎 and 𝑏, expressed as \(𝑃(𝑎<𝑋<𝑏)\).

- Similar definition applies to \(𝑃(𝑋<𝑏)\) and \(𝑃(𝑎<𝑋)\).
- The total probability (the total area under the density curve) equals to 1.

Practice:

1. Find \(𝑃(𝑋<45)\).
   
2. What is the percentage of commuting time that is less than 45 minutes?

Note: \(𝑃(𝑋<45)=𝑃(𝑋≤45)\) – the equal sign does not make difference here in the case of continuous random variable because \(𝑃(𝑋=45)=0\).

Example. Below are a some histograms with different bin widths (10, 5, 2.5, 1.25) for the heights of U.S. adults.

• The estimated probability is \(\frac{195,307+156,239}{3,000,000}=0.1172\)

- For continuous distribution, P(height in [180, 185)) = area under the curve between 180 and 185 =0.1157

Discussion:

1. What does the value of \(\mu\) represent?
   
2. How does the value of \(\sigma\) affect the shape of the curve?

Using the function pnorm():

pnorm(84, mean =72, sd = 15.2)

pnorm(84, mean =72, sd = 15.2, lower.tail = FALSE)

Using the function pnorm():

pnorm(90, mean =72, sd = 15.2)- pnorm(70, mean =72, sd = 15.2)

1 - (pnorm(70, mean =72, sd = 15.2) + pnorm(90, mean =72, sd = 15.2, lower.tail = FALSE))

- The standard normal distribution is \(𝑁(0,1)\). That is, \(\mu=0\) and \(\sigma=1\).
- The letter \(Z\) is used for the random variable that has standard normal distribution.
- The cumulative probability under the standard normal distribution is given by a table (see Textbook, Appendix C, page 410-411) – See an Example on next slide.
- The function pnorm() can also be used to compute probabilities under the standard normal distribution by setting the mean = 0 and sd = 1 or simply not specifying them.
-Example: Suppose \(Z\sim N(0,1)\). Find \(P(Z<1)\).

pnorm(1, mean =0, sd = 1)

pnorm(1)

For \(𝑁(0,1)\), what percent of the variable is in each region? Sketch the graph for each region and use the code chunk below to find the probabilities using pnorm().
1) \(𝑍<1.25\)
2) \(𝑍>−0.25\)
3) \(−0.4 <𝑍<1.5\)
4) \(|𝑍|<1.25\)
5) \(|𝑍|> 2.23\)

For 𝑁(0,1) , what percent of the variable is in each region? Sketch the graph for each region.
1) 𝑍<1.25
2) 𝑍>−0.25
3) −0.4 <𝑍<1.5
4) |𝑍|<1.25
5) |𝑍|> 2.23

- For an observation \(𝑥\) of \(𝑋\), the z-score (standardized scores) is \(𝑧=\frac{𝑥−\mu}{\sigma}\) where \(\mu\) is the mean, \(\sigma\) is the standard deviation

Based on data from smartphones available from major carriers in the U.S, in 2014, the distribution of the standby time approximately follows a normal distribution with a mean of \(\mu=330\) minutes and a standard deviation of \(\sigma=80\) minutes.

As \(\mu=330\), \(\sigma=80\) , we can compute \(\mu−1.25\sigma\), \(\mu+1.25\sigma\):
\(\mu−1.25\sigma=330−1.25×80=230\),
\(\mu +1.25 \sigma=330+1.25×80=430\).

1. 10.56% smartphones have standby time shorter than 230 minutes;
2. 10.56% smartphones have standby time longer than 430 minutes;
3. 78.88% smartphones have standby time between 230 and 430 minutes.

At Heinz ketchup factory the amounts which go into bottles of ketchup are supposed to be normally distributed with mean 36 oz. and standard deviation 0.11 oz. Once every 30 minutes a bottle is selected from the production line, and is below 35.8 oz. or above 36.2 oz., then the bottle fails the quality control inspection.

pnorm(35.8, mean= 36, sd= 0.11) #part (1)

pnorm(36.2, mean= 36, sd= 0.11) - pnorm(35.8, mean= 36, sd= 0.11) #part (2)

We cannot just compare the two raw scores. We instead compare how many standard deviations beyond the mean each score is by using the Z-scores.

• Ann’s score is \(\frac{(1300−1100)}{200}= 1 \ \text{standard deviation above the mean}\)
  
• Tom’s score is \(\frac{24−21}{5}=0.6 \ \text{standard deviations above the mean}\)
  
• Therefore, Ann’s score is better than Tom’s score.

• A \(𝒑^{th}\) percentile is a score below which p% of observations falls below or at the value of the score.

• Graphically, percentile is the score(observation) of random variable that that area (probability)under the probability distribution curve to the left of that observation is 𝑝/100. It is also called the value of the inverse cumulative density function.

For any normal distribution, the mean 𝜇 is the 50th percentile.

For \(N(0,1)\), −1.96 is the 2.5th percentile: the probability that a normal random variable falls at least 1.96 standard deviations below (because of the negative sign) the mean is 0.025, or \(𝑃(𝑍<−1.96)=0.025\)

The function qnorm() returns the value of the \(p(\times 100)th\) percentile under the the normal distribution curve (i.e., the value that has \(p\) probability to its left under the normal curve).

qnorm(0.25)

qnorm(0.5)

qnorm(0.75)

qnorm(0.95, 10, 2)

- For nearly normally distributed data,

Note: Empirical rule is not exactly normal.

Question: For men’s height, within what interval, do the 68% of men’s height fell? 95%? 99.7%?

Exercise. Do the same for women’s height data.

SAT scores are distributed nearly normally with mean 1500 and standard deviation 300.

• \(\sim\) 68% of students score between 1200 and 1800 on the SAT.
• \(\sim\) 95% of students score between 900 and 2100 on the SAT.
• \(\sim\) 99.7% of students score between 600 and 2400 on the SAT.