What is 7/10 in a decimal and percent

Brian can get to class by taking either a Commuter North bus, a Bursley Baits bus, or a Northwood Express bus. If he takes Commu

jolli1 [7]

Answer:

a) P(C/L) = 0.6668

b) P(C/L') = 0.1875

Step-by-step explanation:

Let's call C the event that Brian takes Commuter North, B the event that Brian takes Bursley Baits, N the event that Brian takes Northwood Express, L the event that Brian is late and L' the event that Brian is not late.

First, there is equal probability of taking any given bus so, P(C)=P(B)=P(N)=1/3

Now, the probability P(C/L') that Brian took a Commuter North bus given that he is late is calculated as:

P(C/L) = P(C∩L)/P(L)

Where P(L) = P(C∩L) + P(B∩L) + P(N∩L)

Then, the probability P(C∩L) that Brian takes a Commuter North bus and it is late is calculated as:

P(C∩L)= (1/3)*(0.5) = 0.1667

Because, there is a probability of 1/3 to takes Commuter North Bus and if Brian takes Commuter North there is a 0.5 chance to be late.

At the same way, we get:

P(B∩L) = (1/3)(0.2) = 0.0667

P(N∩L) = (1/3)(0.05) = 0.0167

So, P(L) and P(C/L) are equal to:

P(L) = 0.1667 + 0.0667 + 0.0167 = 0.25

P(C/L) = 0.1667/0.25 = 0.6668

For part b, the probabilities of C, B and N changes and are equal to:

P(C) = 0.3

P(B) = 0.1

P(N) = 0.6

Then, the probability P(C/L') that he took a Commuter North bus given that Brian was not late to class is calculated as:

P(C/L') = P(C∩L')/P(L')

Where P(L') = P(C∩L') + P(B∩L') + P(N∩L')

So, P(C∩L'), P(B∩L') and P(N∩L') are equal to:

P(C∩L') = 0.3*(0.5) = 0.15

P(B∩L') = 0.1*(0.8) = 0.08

P(N∩L') = 0.6*(0.95) = 0.57

It means that P(L') and P(C/L') are equal to:

P(L') = 0.15 + 0.08 + 0.57 = 0.8

P(C/L') = 0.15/0.8 = 0.1875

5 0

3 years ago

Read 2 more answers

2. A marketing firm is trying to estimate the proportion of potential car buyers that would consider

Maurinko [17]

Answer:

a. The number of people that should be in the pilot study are 600 people

b. The point estimate is 0.62 $\overline 6$

c. At 95% confidence level the true population proportion of potential car buyers of hybrid vehicle is between the confidence interval (0.588, 0.6654)

d. Two ways to reduce the margin of error are;

1) Reduce the confidence interval

2) Use a larger sample size

Step-by-step explanation:

a. The given parameters for the estimation of sample size is given as follows;

The margin of error for the confidence interval, E = 4% = 0.04

The confidence level = 95%

The sample size formula for a proportion as obtained from an online source is given as follows;

$n = \dfrac{Z^2 \times P \times (1 - P)}{E^2}$

Where, P is the estimated proportions of the desired statistic, therefore, we have for a new study, P = 0.5;

Z = The level of confidence at 95% = 1.96

n + The sample size

Therefore, we have;

$n = \dfrac{1.96^2 \times 0.5 \times (1 - 0.5)}{0.04^2} = 600.25$

Therefore, the number of people that should be in the pilot study in order to meet this goal at 95% confidence level is n = 600 people

b. The point estimate for the population proportion is the sample proportion given as follows;

$\hat p = \dfrac{x}{n}$

Where;

x = The number of the statistic in the sample

n = The sample size

From the question, we have;

The number of potential car buyers, n = 600

The number of respondent in the sample that indicated that they would consider purchasing a hybrid, x = 376

Therefore, the point estimate, for the proportion of potential car buyers that would consider buying a hybrid vehicle, $\hat p$ = 376/600 = 0.62 $\overline 6$