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3 years ago

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The critical path for a project was found to be 37 days. The variances total 27. What is probability of completing in 43 days? B

elow is information that might be useful to answer the question. Answer to 2 decimal points

1 answer:

Phoenix [80]3 years ago

6 0

Answer:

0.1241.

Step-by-step explanation:

So, from the question we are given the following parameters or information or data which are going to help us in solving this particular question/problem. They are; [1]. The critical path for a project = 37 days, the variances total or variance of critical path or sum of variance on critical path]². = 27, the desired completion time/day = 43 days.

So, we will be making use of the formula given below;

The number of standard deviation = desired completion time/day - critical time or sum of critical path times ÷√[variance of critical path or sum of variance on critical path]².

The number of standard deviation = 43 - 37 ÷ √27 = 1.1547.

Thus, the probability of completing in 43 days = P(Z > 1.1547) = 0.1241.

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Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a

Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(c) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let <em>S</em> = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = <em>p</em> = 0.30.

The number of sample selected is, <em>n</em> = 25.

The random variable $S\sim Bin(25,0.30)$

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion ( $\hat p$ ) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: $E(\hat p)=p=0.30$

The standard deviation of the the sampling distribution of the sample proportion is:

$SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092$

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: $p=\frac{5}{25} =0.20$

$P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

$P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935$

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: $p=\frac{7}{25} =0.28$

$P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

$P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118$

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: $p=\frac{9}{25} =0.36$

$P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

$P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106$

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: $p=\frac{11}{25} =0.44$

$P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

$P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558$

The exact probability is more than the approximated probability.

6 0

3 years ago

I need some help on Slope-Intercepts

serious [3.7K]

The answer is y=3x-4 and here are the steps

6 0

2 years ago

This equation represents an exponential function that passes through the point (2, 80).

miss Akunina [59]

we know that

If the point belongs to the graph, then the point must satisfy the equation

we will proceed to verify each case

<u>case A.)</u> $f(x)=4(x^{5})$

The point is $(2,80)$

Verify if the point satisfy the equation

For $x=2$ find the value of y in the equation and compare with the y-coordinate of the point

$f(2)=4(2^{5})$

$f(2)=128$

$128\neq 80$

therefore

the equation $f(x)=4(x^{5})$ not passes through the point $(2,80)$

<u>case B.)</u> $f(x)=5(x^{4})$

The point is $(2,80)$

Verify if the point satisfy the equation

For $x=2$ find the value of y in the equation and compare with the y-coordinate of the point

$f(2)=5(2^{4})$

$f(2)=80$

$80=80$

therefore

the equation $f(x)=5(x^{4})$ passes through the point $(2,80)$

<u>case C.)</u> $f(x)=4(5^{x})$

The point is $(2,80)$

Verify if the point satisfy the equation

For $x=2$ find the value of y in the equation and compare with the y-coordinate of the point

$f(2)=4(5^{2})$

$f(2)=100$

$100\neq 80$

therefore

the equation $f(x)=4(5^{x})$ not passes through the point $(2,80)$

<u>case D.)</u> $f(x)=5(4^{x})$

The point is $(2,80)$

Verify if the point satisfy the equation

For $x=2$ find the value of y in the equation and compare with the y-coordinate of the point

$f(2)=5(4^{2})$

$f(2)=80$

$80=80$

therefore

the equation $f(x)=5(4^{x})$ passes through the point $(2,80)$

therefore

<u>the answer is</u>

$f(x)=5(x^{4})$

$f(x)=5(4^{x})$

6 0

3 years ago

Read 2 more answers

Help me pleaseeeeeee

Tanzania [10]

-5,7 would be the answer you are looking for.

3 0

3 years ago

Greg's found 72-24 first be subtracted 20 because he think it was easier.Use words and numbers to explain greg's answer

omeli [17]

answer:

its correct as long as he subtracts 4 after that.

Step-by-step explanation:

That is because if you only subtract part of the first number, the answer will be wrong. and the question is also written a bit wierd.

4 0

3 years ago