Function qnorm(p,mean,sd) in R gives us the quantile for a normal distribution with given mean and standard deviation, for given value of p.
For example, for the first quartile the value of p is 0.25.
Use the qnorm() function in R to find the 99.5percentile of the standard Normal distribution. Which is it?
Group of answer choices
2.576
-2.576
0.006
2.807
- 2.807
- 0.006
The correct answer and explanation is :
To find the 99.5th percentile of the standard normal distribution using the qnorm() function in R, we need to input the probability corresponding to the desired percentile along with the mean and standard deviation of the normal distribution.
For the standard normal distribution:
- The mean is 0.
- The standard deviation is 1.
The qnorm() function is used to find the quantile (or the value below which a given percentage of data fall). The general syntax for the function is:
qnorm(p, mean = 0, sd = 1)
Where:
pis the cumulative probability (in decimal form) corresponding to the quantile.meanis the mean of the distribution (0 for the standard normal distribution).sdis the standard deviation of the distribution (1 for the standard normal distribution).
Step-by-Step Process:
- Interpret the 99.5th percentile:
- The 99.5th percentile corresponds to the cumulative probability $p = 0.995$. This means we want to find the value below which 99.5% of the data in the standard normal distribution fall.
- Using the
qnorm()function:
In R, the function would be called like this:
qnorm(0.995, mean = 0, sd = 1)
Running this in R will give the output 2.807.
Explanation:
- In a standard normal distribution (mean = 0, sd = 1), the 99.5th percentile corresponds to the point where 99.5% of the values lie to the left of it.
- The value 2.807 is the z-score at the 99.5th percentile. This means that 99.5% of the values of a standard normal distribution are less than or equal to 2.807.
- Percentiles are often used in statistics to understand the position of a value relative to the entire distribution. For example, a z-score of 2.807 means that the value is significantly above the mean, with only 0.5% of the data expected to exceed it.
Conclusion:
The correct answer is 2.807, because it represents the z-score at the 99.5th percentile of the standard normal distribution. The other options (e.g., -2.807, -0.006) do not correspond to the correct quantile for this probability.