Answer:
Step-by-step explanation:
Assuming there is a punitive removal of one point for an incorrect response.
Five undiscernable choices: 20% chance of guessing correctly -- Expectation: 0.20*(1) + 0.80*(-1) = -0.60
Four undiscernable choices: 25% chance of guessing correctly -- Expectation: 0.25*(1) + 0.75*(-1) = -0.50
I'll use 0.33 as an approzimation for 1/3
Three undiscernable choices: 33% chance of guessing correctly -- Expectation: 0.33*(1) + 0.67*(-1) = -0.33 <== The approximation is a little ugly.
Two undiscernable choices: 50% chance of guessing correctly -- Expectation: 0.50*(1) + 0.50*(-1) = 0.00
And thus we see that only if you can remove three is guessing neutral. There is no time when guessing is advantageous.
One Correct Answer: 100% chance of guessing correctly -- Expectation: 1.00*(1) + 0.00*(-1) = 1.00
Answer:
(a) The expected number of should a salesperson expect until she finds a customer that makes a purchase is 0.9231.
(b) The probability that a salesperson helps 3 customers until she finds the first person to make a purchase is 0.058.
Step-by-step explanation:
Let<em> </em>the random variable <em>X</em> be defined as the number of customers the salesperson assists before a customer makes a purchase.
The probability that a customer makes a purchase is, <em>p</em> = 0.52.
The random variable <em>X</em> follows a Geometric distribution since it describes the distribution of the number of trials before the first success.
The probability mass function of <em>X</em> is:

The expected value of a Geometric distribution is:

(a)
Compute the expected number of should a salesperson expect until she finds a customer that makes a purchase as follows:


This, the expected number of should a salesperson expect until she finds a customer that makes a purchase is 0.9231.
(b)
Compute the probability that a salesperson helps 3 customers until she finds the first person to make a purchase as follows:

Thus, the probability that a salesperson helps 3 customers until she finds the first person to make a purchase is 0.058.
Answer:
a=327 m=416
Step-by-step explanation:
subtract the numbers and add to make sure ur answer is correct
Answer:
x+6<20
x<14
Step-by-step explanation: