I need the answer to these problem...I’ve been stuck on it the most.

A student who is trying to write a paper for a course has a choice of two topics, A and B. If topic A is chosen, the student wil

algol13

Answer:

P(Topic A) = 0.91

P(Topic B) = 0.9163

The student should choose Topic B to maximize the probability of writing a good paper because P(Topic B)>P(Topic A).

Step-by-step explanation:

If Topic A is chosen, 2 books will be ordered (n=2)

If topic B is chosen, 4 books will be ordered (n=4)

Probability that a book arrives in time (p) = 0.7

Probability that a book does not arrive in time (q) = 1 - p = 1 - 0.7 = 0.3

We will <u>use the binomial distribution</u> to find out which topic should the student choose to maximize the probability of writing a good paper. The binomial distribution formula is:

P(X=x) = ⁿCˣ pˣ qⁿ⁻ˣ

where p = probability of success

q = probability of failure

n = total no. of trials

x = no. of successful trials

For Topic A, n = 2, p=0.7 and q=0.3. If topic A is chosen, the student will use at least half the books i.e. he will use either 1 or 2 books. So,

P(Topic A) = P(X=1) + P(X=2)

=²C₁ (0.7)¹(0.3)²⁻¹ + ²C₂ (0.7)²(0.3)²⁻²

= 0.42 + 0.49

P(Topic A) = 0.91

For topic B, n=4, p=0.7 and q=0.3. If topic B is chosen, the student will choose 2 or more books i.e. 2, 3 or 4 books.

P(Topic B) = P(X=2) + P(X=3) + P(X=4)

= ⁴C₂ (0.7)²(0.3)⁴⁻² + ⁴C₃ (0.7)³(0.3)⁴⁻³ + ⁴C₄ (0.7)⁴(0.3)⁴⁻⁴

= 0.2646 + 0.4116 + 0.2401

P(Topic B) = 0.9163

The student should choose Topic B to maximize the probability of writing a good paper because P(Topic B)>P(Topic A) as calculated above.

3 0

3 years ago

As a store manger you would like to find out the average time it takes to unload the truck which delivers the merchandise for yo

Georgia [21]

Answer:

The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes

Step-by-step explanation:

We have the standard deviation for the sample, but not for the population, so we use the students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 35 - 1 = 35

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 34 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.975([tex]t_{975}$ ). So we have T = 2.0322

The margin of error is:

M = T*s = 2.0322*30 = 60.97

The upper end of the interval is the sample mean added to M. So it is 204 + 60.97 = 264.97

The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes

4 0

3 years ago

Read 2 more answers

A table with four legs will sometimes wobble if one leg is shorter than the other three, but a table with three legs will not wo

prohojiy [21]

Because it is made to balance like that.

5 0

2 years ago

The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and me

Dovator [93]

Answer:

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

$e = 2.71828$ is the Euler number

$\mu$ is the mean in the given time interval.

The problem states that:

The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and mean 0.2 each day during the weekend.

To find the mean during the time interval, we have to find the weighed mean of calls he receives per day.

There are 5 weekdays, with a mean of 0.1 calls per day.

The weekend is 2 days long, with a mean of 0.2 calls per day.

So:

$\mu = \frac{5(0.1) + 2(0.2)}{7} = 0.1286$