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The conditional probability that an adult who lifts weights regularly, also runs regularly is 0.219 or 21.9% if the 18% lift weights regularly, 32% run regularly, and 7% lift weights and run regularly.
<h3>What is probability?</h3>It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words the probability is the number that shows the happening of the event.
Let's suppose P(A) is the probability of adults who lift weights regularly.
P(A) = 18% = 0.18
Similarly,
P(B) = 32% = 0.32
P(A∩B) = 0.07
We know:
P(A|B) = 0.219 or
P(A|B) = 21.9%
Thus, the conditional probability that an adult who lifts weights regularly, also runs regularly is 0.219 or 21.9% if the 18% lift weights regularly, 32% run regularly, and 7% lift weights and run regularly.
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Answer:
Step-by-step explanation:
From the given question;
Given that:
Jan's All You Can Eat Restaurant charges $9.10 per customer to eat at the restaurant.
Distribution is skewed and and has a mean of $8.10 and a standard deviation of $4.
A. If the 100 customers on a day have the characteristics of the random sample from their customer base, find the mean and standard error of the sampling distribution of the restaurant's sample mean expense per customer.
the mean by using the central limit theorem is 8.10
the standard error of the sampling distribution =
the standard error of the sampling distribution =
= 4/10
= 0.4
B.
P(X > $8.95) = P (Z > 8.95 - 8.10/0.4)
P(X > $8.95) = P (Z > 2.1)
P(X > $8.95) = 1 - P (Z < 2.1)
P(X > $8.95) = 1 - 0.9821
P(X > $8.95) = 0.0179