I'm given a sample data of heights of customers for a coffee shop in previous week.
y = (1.78,1.65,1.62,1.84,1.75,1.85,1.52,1.55)
The mean of the dataset is 1.695 and the standard deviation is 0.1279509.
I've used R to generate a few probabilities for certain scenarios:
For customers of height $<1.5$m,
P(X<1.5) = 0.05163023
For customers of height $>1.8$m,
P(X>1.8) = 0.1901652
For customers of height $> 1.5$m and $< 1.65$m,
P(1.5 < X < 1.65) = 0.3018355
For customers of height > 1.65m and < 1.8m,
P(1.65 < X < 1.8) = 0.456369
I have a question which I couldn't seem to figure out.
"If the coffee shop receives 10 customers in one day, what is the probability that none of them will be 1.65m or taller."
I'm given a sample data of customers in previous week but the question states "10 customers in one day". I was wondering how would we calculate the probability if so?
I attempted something like:
P(X>= 1.65) = 0.6465343
But it doesnt seem like I'm using the "10" as the question stated, and the probability above is based on the data for previous week not on the day.
Would like some help on this.
Best Answer
So, your answer $\mathbb{P}(X \ge 1.65) \approx 0.65$ is correct, however this indicates the probability of one costumer to be $1.65$m or taller. To answer your original question, let $X_1, \dots, X_{10}$ be the heights of customers and the probability of our interest is $$ \mathbb{P}(X_{1} < 1.65, \dots, X_{10} < 1.65) = \mathbb{P}(X_1 < 1.65)^{10} \approx 0.35^{10} \approx 0.0000276, $$ assuming independence of these customers.