- Also known as the Z distribution
- Mean is 0
- Standard deviation is 1
- Characteristics of a Normal Distribution
- Continuous Random Variable
- - ∞ < x < + ∞
- Curve is symmetrical around the mean (m).
- Area under curve = 1
- Mean & standard deviation uniquely determine a normal distribution.
To find P(a < X < b) when X is distributed normally:
- Draw the normal curve for the problem in terms of X
- Translate X-values to Z-values and put Z values on your diagram
- Use the Standardised Normal Table
Example: Suppose X is normally distributed with mean 8 and std dev 5. Find P(X < 8.6)
Finding the X for a Known Probability:
- Draw a normal curve placing all known values on it such as mean of X and Z
- Shade in area of interest and find cumulative probability
- Find the Z value for the known probability
- Convert to X units using the formula:
How Large is Large Enough?
- For most population distributions, n ≥ 30 will give a sampling distribution that is nearly normal
- For fairly symmetric population distributions, n ≥ 5 is sufficient
- For normal population distributions, the sampling distribution of the mean is always normally distributed
- A point estimate is the value of a single sample statistic
- A confidence interval provides a range of values constructed around the point estimate
Confidence Level (1-a)
- Common confidence levels = 90%, 95% or 99%
- Also written (1 – a) = 0.90, 0.95 or 0.99
- A relative frequency interpretation
- In the long run, 90%, 95% or 99% of all the confidence intervals that can be constructed (in repeated samples) will contain the unknown true parameter
- For example, if we were to randomly select 100 samples and use the results of each sample to construct 95% confidence intervals, approximately 95 out of 100 would contain the population mean
So, what happens if we don’t know the standard deviation of the population????
- If the population standard deviation σ is unknown, we can substitute the sample standard deviation, S
- This introduces extra uncertainty, since S is variable from sample to sample
- So we use the t distribution instead of the normal distribution
Confidence Interval Estimate:
where t is the critical value of the t distribution with n -1 degrees of freedom and an area of α/2 in each tail
A random sample of n = 25 has X = 50 & S = 8.
Form a 95% confidence interval for μ:
d.f. = n – 1 = 24, so
The confidence interval is 46.698 ≤ μ ≤ 53.302
Required Sample Size Example
If s = 45, what sample size is needed to estimate the mean within ± 5 with 90% confidence?
So the required sample size is n = 220