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A real estate agent has 12 properties that she shows. She feels that there is a 30% chance of selling any one property during a

Ronch [10]

Answer:

0.2528 = 25.28% probability of selling no more than 2 properties in one week.

Step-by-step explanation:

For each property, there are only two possible outcomes. Either they are sold, or they are not. The chance of selling any one property is independent of selling another property, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A real estate agent has 12 properties that she shows.

This means that $n = 12$

She feels that there is a 30% chance of selling any one property during a week.

This means that $p = 0.3$

Compute the probability of selling no more than 2 properties in one week.

2 or less sold, which is:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.3)^{0}.(0.7)^{12} = 0.0138$

$P(X = 1) = C_{12,1}.(0.3)^{1}.(0.7)^{11} = 0.0712$

$P(X = 2) = C_{12,2}.(0.3)^{2}.(0.7)^{10} = 0.1678$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0138 + 0.0712 + 0.1678 = 0.2528$

0.2528 = 25.28% probability of selling no more than 2 properties in one week.

8 0

3 years ago

3. Last year, the numbers of calculators produced per day at a certain factory were normally distributed with a mean of 560 calc

Mademuasel [1]

Answer:

3. Last year, the numbers of calculators produced per day at a certain factory were normally distributed with a mean of 560 calculators and a standard deviation of 12 calculators.(C)

Step-by-step explanation:

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8 0

3 years ago

FIRST PERSON TO ANSWER GETS BRAINLIEST OR FIVE STARS Define the domain and range of the function.

uysha [10]

Answer:

Domain: (-∞, ∞)

Range: (-∞, ∞)

Step-by-step explanation:

The domain are the x-values included in the function (the horizontal axis).

The range are the y-values included in the function (the vertical axis).

The two arrows on the ends of the line (pointing upwards and downwards respectively) indicate that the function goes in those direction for infinity. Therefore, if there are an infinite amount of y-values, the range is (-∞, ∞).

While the slope is quite steep, there is still a slope and slowly "expands" the line on the horizontal axis. Because there is no limit to the y-values, the domain will also expand infinitely. Therefore, the domain is also (-∞, ∞).

7 0

2 years ago

Read 2 more answers

(Easy) For the following quadratic function, find the axis of symmetry, the vertex and the y-intercept. y = x^2 + 12x + 32

hammer [34]

Answer:

$y ( \times + 6) 2 - 4$

7 0

3 years ago

Read 2 more answers

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0

3 years ago