Understanding the Normal Distribution and Calculating X Values

When dealing with normal distributions, it is often useful to determine specific values of X given a certain probability. In the scenario presented, we are given a normal distribution with an average (mean, mu) of 70 and a standard deviation (sigma) of 10. The task is to find the value of X such that the probability of X being less than or equal to a certain value is 91%, or P(X leq a) 0.91.

Step-by-Step Calculation

The first step is to recognize that we need to convert this probability into a z-score. The z-score for an area of 0.09 (1 - 0.91) under the normal curve can be found using statistical tables or software tools. In R, this can be done using:

library(tigerstats) qnorm(0.09)

The result will be approximately -1.34. This z-score corresponds to the left tail of the normal distribution.

Using the Standard-Score Formula

The standard-score (z-score) formula is given by:

z frac{x - mu}{sigma}

Manipulating this formula to solve for x:

Start with: z frac{x - mu}{sigma} Multiply both sides by sigma z cdot sigma x - mu Add mu to both sides x z cdot sigma mu

Substituting the values we know (z -1.34, sigma 10, and mu 70):

x (-1.34) cdot 10 70 -13.4 70 56.6

Therefore, the value of X that satisfies P(X leq a) 0.91 is 56.6.

Verification Using a TI-84 Calculator

For those using a TI-84 calculator, the inverse normal cumulative distribution function (invNorm) can be used for verification:

Press 2ND VARS 3 for invNorm. Enter: invNorm(0.91, 70, 10). Press ENTER to calculate.

The result should be approximately 56.59244965, which confirms our manual calculation.

Estimate and Verification of z-Score

Before performing the calculation, it can be helpful to estimate the z-score based on a rough sketch of the normal curve.

Given the mean (mu) of 70 and a standard deviation (sigma) of 10, we can mark the curve with the following divisions:

2 standard deviations left: 50 1 standard deviation left: 60 Mean: 70 1 standard deviation right: 80 2 standard deviations right: 90 3 standard deviations right: 100

Since 98% of the data lies within 2.33 standard deviations below the mean, we can estimate that the value is between 50 and 60. Using the invNorm function on the calculator (invNorm(0.91, 70, 10)) confirms the precise value.

Conclusion

By understanding and utilizing the properties of the normal distribution and the z-score, we can accurately determine specific values of X based on given probabilities. This method is not only useful in statistical analysis but also in numerous practical applications in science, engineering, and business.