I was watching this video from Khan Academy and the question was to find the probability of a random variable (with a normal distribution) being less than a certain number. The instructor finds the answer by using a Z table. My question is since he found the probability of a random variable being less than a number, how would I find the probability of a random variable being less than or equal to a number?
Finding the probability of a random variable (with a normal distribution) being less than or equal to a number using a Z table
probabilityprobability distributionsprobability theoryrandom variablesstatistics
Related Solutions
If the discrete random variable $X$ takes integer values, then $$P(X > x)= P(X \ge x+1) = P(X \ge x+.5)$$ The continuity correction would use the third expression when using a continuous distribution as an approximation.
Ordinarily, the approximating continuous distribution would have positive probability in the interval $[x, x+.5].$ In that case using the continuity correction will give you a smaller approximated value.
Example: Suppose $X \sim \mathsf{Binom}(n = 64, p = 1/2)$ and you seek $P(X > 30).$ The exact value is $P(X > 30) = 1 - P(X \le 30) = 0.6460096.$
1 - pbinom(30, 64, .5)
## 0.6460096
If you use $P(X^\prime > 30) = 1 - P(X^\prime \le 30)$ as an approximation, where $X^\prime \sim \mathsf{Norm}(\mu = 32, \sigma=4),$ you will get $P(X > 300) \approx 0.6914625.$
1 - pnorm(30, 32, 4)
## 0.6914625
But if you use the continuity correction, you will use $P(X^\prime > 30.5) = 1 - P(X^\prime \le 30.5) = 0.6461698.$ Hence, your approximation will be $P(X > 30) \approx 0.6461698.$ This is smaller than the value 0.6914625 without the continuity correction. It is also closer to the exact binomial probability.
1 - pnorm(30.5, 32, 4)
## 0.6461698
Usually in textbook examples you can expect about two decimal places of accuracy from a continuity-corrected normal approximation to a binomial distribution. To four decimal places, the exact value in this example is 0.6460 and the continuity-corrected normal approximation is 0.6462. (Here we get three-place accuracy; approximations are often best when $p \approx 1/2.$)
The figure below shows relevant binomial probabilities (vertical bars) and the approximating normal density curve. Notice that be binomial probability $P(X = 31)$ is approximated by the area under the normal curve above the interval $[30.5, 31.5].$ The uncorrected approximation wrongly includes the vertical strip between $x = 30.0$ and $x=30.5$ under the normal curve.
Note: The values I have shown are from R statistical software. If your normal approximations are obtained by standardization and using a printed normal table, then results will be slightly different because of the rounding entailed in the use of the table.
I had the very same issue. As @mihaild mentions in the comment, your solution is the inverse of the true one. The real reason for this is in the text though.
What you (and I) calculated was the probability that Yuki's time is greater than Zana's. However, the question is about the probability that Yuki's time is faster, not greater. Because, of course, she is faster when her time is lower.
Best Answer
Since the normal distribution is a continuous distribution, \begin{align*} \mathcal{P}(X \leq c) &= \mathcal{P}((X < c) \cup (X = c)) \\ &= \mathcal{P}(X < c) + \mathcal{P}(X = c) - \mathcal{P}((X < c) \cap (X = c)) \\ &= \mathcal{P}(X < c) + 0 - 0 \\ &= \mathcal{P}(X < c) \text{.} \end{align*} The first zero, $\mathcal{P}(X = c) = 0$, is because the normal distribution is a continuous distribution : probabilities are given by integrals of its probability density function, but we are asking for the integral $\int_c^c \dots = 0$ which is zero because the endpoints of integration are the same. The second zero, $\mathcal{P}((X < c) \cap (X = c)) = 0$, is zero because there are no realizations of $X$ that are simultaneously less than $c$ and equal to $c$.
So, the two probabilities you ask about are the same. Use the same procedure for both.