1. For a group of men, the probability distribution for the number x who live through the next year is as given in the accompanying table. Find its mean and standard deviation. 2.... 1. For a group of men, the probability distribution for the number x who live through the next year is as given in the accompanying table. Find its mean and standard deviation. 2. In a study of brand recognition of Samsung, groups of five consumers are interviewed. If x is the number of people in the group who recognize the Samsung brand name, then x can be 0, 1, 2, 3, 4 or 5 and the corresponding probabilities are 0.0016, 0.1250, 0.0432, 0.2982, 0.0078 and 0.4096. Is x a probability distribution? 3. When you give a casino $5 for a bet on the pass line in a casino game of dice, there is a 251/495 probabilities that you will lose $5 and there is 244/495 probabilities that you will make a net gain of $5. (If you win, casino gives you $5 and you get to keep your $5 bet, so the net gain is $5). What is your expected value? In the long run, how much do you lose for each dollar bet? 4. Let the random variable x represent the number of girls in a family of four children. Construct a table describing the list of probability distribution, then find the mean and standard deviation. (Hint: list the difference possible outcomes.) Is it unusual for a family of four children to consist of four girls? 5. The author purchased a slot machine that is configured so that there is a 1/2000 probability of winning any jackpot on any individual trial. Although no one would seriously consider tricking the author, suppose that a guess claims that she played the slot machine 5 times and hit the jackpot twice. (a) Find the probability of exactly two jackpots in 5 trials. (b) Find the probability of at least two jackpots in 5 trials. (c) Does the guest's claim of two jackpots in 5 trials seem valid? Explain. 6. In a test of the MicroSort method of gender selection, 51 babies are born to couples trying to have baby boys and 39 of those babies are boys. (a) If the gender selection method has no effect and boys and girls are equally likely, find the mean and standard deviation for the numbers of boys born in group of 51. (b) Is the result of 39 boys unusual? 7. The author found that in one month (30 days), he made 47 cell phone calls, which were distributed as follows: No calls were made on 17 days, 1 call was made on each of 7 days, 3 calls were made on each of two days, 12 calls were made on one day and 14 were made on one day.
Understand the Problem
The question presents a series of problems related to probability distributions and asks for calculations such as mean, standard deviation, and the probabilities of certain events occurring. Each question requires applying statistical concepts to analyze different scenarios.
Answer
Mean: $E(X) = 3.9609$, Standard Deviation: $SD(X) \approx 0.191$.
Answer for screen readers
The mean is $E(X) = 3.9609$, and the standard deviation is approximately $SD(X) \approx 0.191$.
Steps to Solve
- Finding the Mean of the Distribution
The mean (expected value) of a probability distribution is calculated using the formula:
$$ E(X) = \sum (x \cdot P(x)) $$
Using the values from the table:
- For $x = 0$: $0 \cdot 0.0000 = 0$
- For $x = 1$: $1 \cdot 0.0001 = 0.0001$
- For $x = 2$: $2 \cdot 0.0006 = 0.0012$
- For $x = 3$: $3 \cdot 0.0387 = 0.1161$
- For $x = 4$: $4 \cdot 0.9609 = 3.8436$
Summing these values gives:
$$ E(X) = 0 + 0.0001 + 0.0012 + 0.1161 + 3.8436 = 3.9609 $$
- Finding the Variance and Standard Deviation
Variance is calculated using:
$$ Var(X) = E(X^2) - (E(X))^2 $$
First, compute $E(X^2)$:
- For $x = 0$: $0^2 \cdot 0.0000 = 0$
- For $x = 1$: $1^2 \cdot 0.0001 = 0.0001$
- For $x = 2$: $2^2 \cdot 0.0006 = 0.0024$
- For $x = 3$: $3^2 \cdot 0.0387 = 0.3483$
- For $x = 4$: $4^2 \cdot 0.9609 = 15.3744$
Thus,
$$ E(X^2) = 0 + 0.0001 + 0.0024 + 0.3483 + 15.3744 = 15.7252 $$
Now substitute into the variance formula:
$$ Var(X) = 15.7252 - (3.9609)^2 $$
Calculating $(3.9609)^2$ gives:
$$ (3.9609)^2 = 15.6887 $$
Finally:
$$ Var(X) = 15.7252 - 15.6887 = 0.0365 $$
The standard deviation is the square root of variance:
$$ SD(X) = \sqrt{Var(X)} = \sqrt{0.0365} \approx 0.191 $$
The mean is $E(X) = 3.9609$, and the standard deviation is approximately $SD(X) \approx 0.191$.
More Information
In probability distributions, the mean represents the expected outcome, while the standard deviation reflects the dispersion of the outcome around the mean. These calculations help in understanding the likelihood of various outcomes in the given scenarios.
Tips
- Forgetting to multiply each $x$ by its corresponding probability $P(x)$ when calculating the mean.
- Confusing variance with standard deviation; variance is the square of the standard deviation.
- Miscalculating the expected value for $E(X^2)$.
AI-generated content may contain errors. Please verify critical information
