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Llana [10]
2 years ago
7

he accompanying data provide the heights of some singers in a​ chorus, collected so that the singers could be positioned on stag

e with shorter ones in front and taller ones in back. The histogram displays these data. Complete parts a and b below. Click here to view the data. LOADING... Click here to view the histogram. LOADING... ​a) Describe the distribution. A. The distribution is bimodal. Its modes are near 65 inches and 71 inches. B. The distribution has no noticeable modes. C. The distribution is unimodal. Its mode is near 60 inches. D. The distribution is multimodal. Its modes are near 60​ inches, 65​ inches, and 71 inches.

Mathematics
1 answer:
stellarik [79]2 years ago
3 0

The image of the histogram distribution is missing, so i have attached it.

Answer:

Option C: The distribution is unimodal. Its mode is near 60 inches

Step-by-step explanation:

Mode simply means the most occurring value in a set of data.

Now, for histogram, the mode is the highest peak represented in the bars.

Now, from the image attached, it is seen that there is a clear highest peak. Thus, the histogram has only one peak and it can be described as unimodal.

We are not given the mode height from the histogram but we can see that it is between 60 and 68 inches. And if carefully observed it will have a value closer to 60.

Thus,option C is correct.

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Answer:

a) P(X ≤ 2) = 0.87

b)  P(X ≥ 5) = 0.01

c) P(1 ≤ X ≤ 4) = 0.71

d) P ( X = 0 ) = 0.28

e) σ(X) = 1.09 , E(X) = 1.25

Step-by-step explanation:

Given:

- Let X = the number of defective boards in a random sample of size n = 25, so X ~ Bin(25, 0.05)

Where, n = 25 and p = 0.05

Find:

(a) Determine P(X ≤ 2).(b) Determine P(X ≥ 5).(c) Determine P(1 ≤ X ≤ 4).(d) What is the probability that none of the 25 boards is defective?(e) Calculate the expected value and standard deviation of X.

Solution:

- The probability mass function for a binomial distribution is given by:

                         P ( X = x ) = nCr * (p)^r * ( 1 - p )^(n-r)

a) P(X ≤ 2):

                         P(X ≤ 2) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 )

                         = (0.95)^25 + 25*0.05*0.95^24 + 25C2*0.05^2*0.95^23

                         = 0.87

b) P(X ≥ 5):

            P(X ≥ 5) = 1 - [P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P(X=4)

            = 1 - [ 0.87 + 25C3*0.05^3*0.95^22 + 25C4*0.05^4*0.95^21]

            = 1 - 0.98994

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c) P(1 ≤ X ≤ 4):

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            P ( X = 0 ) = 0.95^25 = 0.28

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            E(X) = 25*0.05 = 1.25

            σ(X) = sqrt ( Var (X) )

            σ(X) = sqrt ( n*p*(1-p) ) = sqrt ( 25*0.05*0.95 )

            σ(X) = 1.09

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