If the hourly sales at a certain register in a supermarket are normally distributed with a mean of $650 and a standard deviation of $60, find the probability that the sales during a randomly selected hour are greater than $800.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score above.
Ah, sales at a supermarket, a classic topic! Let me compute the probability for you while tickling your funny bone.
First, we need to standardize the problem. We'll use the z-score formula: z = (X - μ) / σ
Where:
X is the value we're interested in - in this case, $800
μ is the mean - in this case, $650
σ is the standard deviation - in this case, $60
Plugging in the values, we get:
z = (800 - 650) / 60
Calculating this, we find that the z-score is approximately 2.5.
Now, we need to find the probability associated with this z-score. To do this, we consult the z-table, also known as the "Table of Z-Scores to Take a Break and Chuckle".
Looking up 2.5 in the table, I realize that I don't need a table because it's one of my favorite jokes! Why did the z-score go to therapy? Because it had standard deviation issues! 🤣
Ahem, now back to business. The probability associated with a z-score of 2.5 is about 0.9938.
Therefore, the probability that the sales during a randomly selected hour are greater than $800 is approximately 0.9938, or about 99.38%.
Remember, probabilities can be unpredictable, just like unexpected items in the bagging area!
To find the probability that the sales during a randomly selected hour are greater than $800, we need to standardize the value of $800 using the formula for standardization:
Z = (X - μ) / σ
where:
Z is the standardized score,
X is the value we want to standardize ($800 in this case),
μ is the mean of the distribution ($650 in this case), and
σ is the standard deviation of the distribution ($60 in this case).
Substituting the given values, we have:
Z = (800 - 650) / 60
Z = 150 / 60
Z ≈ 2.5
Now, we need to find the probability corresponding to a Z-score of 2.5 in the standard normal distribution (where the mean is 0 and the standard deviation is 1). We can use a standard normal distribution table or calculator to find this probability.
The probability can be looked up as P(Z > 2.5) = 1 - P(Z ≤ 2.5).
Using a standard normal distribution table, we find that the probability corresponding to a Z-score of 2.5 is approximately 0.9938. Therefore, the probability that the sales during a randomly selected hour are greater than $800 is approximately 0.9938 (or 99.38%).
To find the probability that the sales during a randomly selected hour are greater than $800, we need to standardize the value of $800 and find the corresponding z-score.
1. Calculate the z-score:
- The z-score formula is: z = (x - μ) / σ
- where x is the value we want to standardize ($800 in this case)
- μ is the mean ($650)
- σ is the standard deviation ($60)
Plugging in the values, we get:
z = (800 - 650) / 60 = 2.5
2. Once we have the z-score, we can use a standard normal distribution table or a calculator to find the probability associated with that z-score.
Using a standard normal distribution table, we look up the probability corresponding to a z-score of 2.5. This probability represents the area under the normal curve to the right of the z-score.
If you don't have access to a standard normal distribution table, you can use an online calculator or software to find the probability directly. For example, using an online calculator, we find that the probability of a z-score of 2.5 is approximately 0.9938.
Therefore, the probability that the sales during a randomly selected hour are greater than $800 is approximately 0.9938 or 99.38%.