According to a recent survey, 75% of all customers will return to the same grocery store. Suppose 8 customers are selected at random:
A) What is the probability that exactly five of the customers will return?
B) What is the profability that all 8 will return?
C) What is the probability that at least on will return?
D) How many customers would be expected to return to the same store?
E) Calculate the variance in the number of customers returning to the same store?
F) Determine the standard Deviation.
According to recent survey 75% of all customers will return yo the same grocery store. Suppose eight customers are selected at random, what
is the probability that:
exactly five of the customers will return?
all eight will return?
At least seven will return?
At least one will returns?
How many customers would be expected to return to the same store?
1. A researcher is interested in whether students who attend private high schools have higher average SAT Scores than students in the general population. A random sample of 90 students at a private high school is tested and and a mean SAT score of 1030 is obtained. The average score for public high school student is 1000 (ó= 200). Is this a one or two tail test and why?
The key to determining one-tailed or two-tailed tests is in the problem. Note the first sentence. This tells you the test is one-tailed because the word "higher" will translate to "greater than" in the alternate hypothesis. One-tailed tests show a specific direction. If the problem was just looking for a difference in either tail, then the test would be two-tailed.
I hope this will help.
A) Well, if exactly five customers out of eight will return, it's like getting five out of eight right on a multiple-choice test. So, the probability would be 5/8, or about 62.5%. That's like hitting the jackpot but with slightly fewer zeros.
B) Now, if you want all eight customers to return, that's a bit of a stretch. It's like hoping for a unicorn to deliver your groceries. The probability of all eight returning would be (0.75)^8, which is approximately 1.98%. So, you're more likely to find Atlantis than to have all your customers return to the same store.
C) The probability that at least one customer will return is the opposite of the probability that none of them will return. Since 100% - 1.98% = 98.02%, we can conclude that there's a whopping 98.02% chance that at least one customer will return. That's almost as guaranteed as finding socks that have disappeared somewhere in your dryer!
D) To calculate the expected number of customers who will return, we simply multiply the probability of each customer returning (0.75) by the total number of customers (8). So, 0.75 * 8 = 6 customers are expected to return. Don't worry, they won't show up all at once, wearing party hats and singing show tunes.
E) To calculate variance, we need to multiply the probability of each outcome (returning or not) by the squared difference between that outcome and the expected value which we just found (6). For returning customers, it's (0.75 - 6)^2 * (8), and for non-returning customers, it's (0.25 - 6)^2 * (8). Add them up, and you get a variance of approximately 9.12. Variance tells you how spread out the data is, kind of like how the grocery carts sometimes seem to have a mind of their own and end up scattered everywhere.
F) The standard deviation is the square root of the variance. So, the square root of 9.12 is approximately 3.02. This tells us how much the number of customers returning tends to deviate from the expected value of 6. It's like how one minute you're expecting a customer to show up, and the next minute you're being chased by a runaway shopping cart.
To answer these questions, we can use the binomial probability formula:
P(x) = (nCx) * p^x * (1-p)^(n-x)
where:
P(x) is the probability of getting exactly x successes,
nCx is the combination formula for choosing x items from a set of n items,
p is the probability of success (in this case, the probability that a customer will return),
(1-p) is the probability of failure (in this case, the probability that a customer will not return),
n is the total number of trials (in this case, the number of customers selected), and
x is the number of successes (in this case, the number of customers who will return).
Let's calculate each part separately:
A) What is the probability that exactly five of the customers will return?
For this part, we need to calculate P(x=5).
n = 8 (total number of customers selected)
x = 5 (number of customers who will return)
p = 0.75 (probability that a customer will return)
Using the formula, we get:
P(x=5) = (8C5) * (0.75)^5 * (1-0.75)^(8-5)
= (8C5) * (0.75)^5 * (0.25)^3
To calculate (8C5), we can use the combination formula:
(8C5) = (8!)/(5!(8-5)!)
= (8!)/(5!3!)
= (8 * 7 * 6)/(3 * 2 * 1)
= 56
Therefore,
P(x=5) = 56 * (0.75)^5 * (0.25)^3
B) What is the probability that all 8 will return?
For this part, we need to calculate P(x=8).
n = 8
x = 8
p = 0.75
Using the formula, we get:
P(x=8) = (8C8) * (0.75)^8 * (1-0.75)^(8-8)
= (8C8) * (0.75)^8 * (0.25)^0
= 1 * (0.75)^8 * 1
Therefore,
P(x=8) = (0.75)^8
C) What is the probability that at least one will return?
For this part, we need to find the probability of getting at least one success, which is the complement of getting zero successes.
P(at least one) = 1 - P(x=0)
n = 8
x = 0
p = 0.75
Using the formula, we get:
P(x=0) = (8C0) * (0.75)^0 * (1-0.75)^(8-0)
= (8C0) * (0.75)^0 * (0.25)^8
= 1 * 1 * (0.25)^8
Therefore,
P(at least one) = 1 - (0.25)^8
D) How many customers would be expected to return to the same store?
To find the expected value, we multiply the total number of customers selected (n) by the probability of success (p).
n = 8
p = 0.75
Expected value = n * p
= 8 * 0.75
E) Calculate the variance in the number of customers returning to the same store?
The variance can be calculated using the formula:
Variance = n * p * (1 - p)
n = 8
p = 0.75
Variance = 8 * 0.75 * (1 - 0.75)
F) Determine the standard deviation.
The standard deviation is the square root of the variance.
Standard deviation = √(Variance)