A project has a mean completion time of 54 weeks and a standard deviation of the project completion time of 2.95 weeks. What is the probability of completing the project within 52 weeks? Please put your answer in terms of a probability (i.e. a number between 0 and 1), and not a percentage.

Respuesta :

Answer:

0.2483

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 54, \sigma = 2.95[/tex]

What is the probability of completing the project within 52 weeks?

This is the pvalue of Z when X = 52.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{52 - 54}{2.95}[/tex]

[tex]Z = -0.68[/tex]

[tex]Z = -0.68[/tex] has a pvalue of 0.2483

So the answer is 0.2483

Answer:

Probability of completing the project within 52 weeks is 0.2483.

Step-by-step explanation:

We are given that a project has a mean completion time of 54 weeks and a standard deviation of the project completion time of 2.95 weeks.

Let X = completion time of  a project

The z-score probability distribution is given by;

                   Z = [tex]\frac{ X -\mu}{{\sigma}} }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = mean completion time = 54 weeks

           [tex]\sigma[/tex] = standard deviation = 2.95 weeks

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability of completing the project within 52 week is given by = P(X [tex]\leq[/tex] 52 weeks)

     P(X [tex]\leq[/tex] 52) = P( [tex]\frac{ X -\mu}{\sigma}} }[/tex] [tex]\leq[/tex] [tex]\frac{52-54}{2.95}} }[/tex] ) = P(Z [tex]\leq[/tex] -0.68) = 1 - P(Z < 0.68)

                                                    = 1 - 0.75175 = 0.2483

Because in z table area of P(Z < x) is given. Also, the above probability is calculated using z table by looking at value of x = 0.77 in the z table which have an area of 0.75175.

Therefore, probability of completing the project within 52 weeks is 0.2483.

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