Statistics Help: Answers to Questions & Explanations

[PARAJON]

I need help on the following Questions that I'm working on. Can someone explain to me and show me how they came up with the answers. Thank you.




#1

The events X and Y are mutually exclusive. Suppose P(X) = .05 and P(Y) = .02. What is the probability of either X or Y occurring? What is the probability that neither X nor Y will happen?


#2

If you ask three strangers on campus, what is the probability: (a) All were born on Wednesday? (b) All were born on different days of the week? (c) None were born on a Saturday?

#3

In a binomial situation n=5 and pie = .40 Determine the probabilities of the following events using the binomial formula.

a. x = 1

b. x = 2


#4

Steele Electronics, Inc. sells expensive brands of stereo equipment in several shopping malls throughout the northwest section of the United States. The Marketing Research Department of Steele reports that 30 percent of the customers entering the store that indicate they are browsing will, in the end, make a purchase. Let the last 20 customers who enter the store be a sample.

a. How many of these customers would you expect to make a purchase

b. What is the probability that exactly five of these customers make a purchase?

c. What is the probability ten or more make a purchase?

d. Does it seem likely at least one will make a purchase?



#5

A recent article in the Myrtle Beach Sun Times reported that the mean labor cost to repair a color TV is $90 with a standard deviation of $22. Monte’s TV Sales and Service completed repairs on two sets this morning. The labor cost for the first was $75 and it was $100 for the second. Compute z values for each and comment on your findings.


#6

The mean starting salary for college graduates in the spring of 2000 was $31,280. Assume that the distribution of starting salaries follows the normal distribution with a standard deviation of $3,300. What percent of the graduates have starting salaries:

a. Between $30,000 and $35,0000

b. More than $40,000

c. Between $35,000 and $40,000

 

[matt grime]

1. Look up the definitoins.
2. think about it a little
3. what's x?
4. see 2.
5. what's z?
6. use the formulae even though the assumptions are ridiculous.


sorry, but the questions that make sense are failrly easy, and this is obviously homework, and a lot more good will come if you try ther homework yourself, perhaps seying what you've done will demonstrate you've tried it at least in part.

 

[tacman]

Hello! I would be happy to help you with these questions and explain how to find the answers.

#1 Probability of either X or Y occurring: Since X and Y are mutually exclusive, they cannot occur at the same time. Therefore, the probability of either X or Y occurring is the sum of their individual probabilities, which is 0.05 + 0.02 = 0.07.

Probability that neither X nor Y will happen: Since they are mutually exclusive, if one occurs, the other cannot occur. Therefore, the probability that neither X nor Y will happen is 1 - probability of either X or Y occurring, which is 1 - 0.07 = 0.93.

#2
(a) Probability that all three were born on Wednesday: Assuming each day of the week is equally likely, the probability of being born on a Wednesday is 1/7. Since the events are independent, we can multiply the probabilities together, so the probability of all three being born on a Wednesday is (1/7)^3 = 0.0029.

(b) Probability that all were born on different days of the week: This is similar to part (a), but now we need to account for the fact that each person can be born on any day except the one(s) already chosen. So, the probability is (1/7) * (6/7) * (5/7) = 0.1429.

(c) Probability that none were born on a Saturday: This is the same as the probability that all were born on a day other than Saturday, which is (6/7)^3 = 0.684.

#3
a. Probability of x = 1: Using the binomial formula, we have P(x=1) = (5 choose 1) * (0.4)^1 * (0.6)^4 = 0.2304.

b. Probability of x = 2: Similarly, P(x=2) = (5 choose 2) * (0.4)^2 * (0.6)^3 = 0.3456.

#4
a. Expected number of customers making a purchase: Since 30% of the customers who enter the store make a purchase, we can expect 30% of the 20 customers to make a purchase, which is 0.30 * 20 = 6 customers.

b. Probability that