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Asked: 2022-03-14T09:41:01+05:30 In:

Question 1 2 Points Assume that on an average one telephone number out of fifteen is busy. What is the …

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Question 1
2 Points
Assume that on an average one telephone number out of fifteen is busy. What is the probability that if six randomly selected telephone numbers are called:

(1) not more than three will be busy?

(2) at least three of them will be busy ?

Question 2
2 Points
Out of 800 families with 4 children each, how many families would be expected to have (1) two boys 2 girls

(2) at least one boy

(3) no girl

(4) at most two girls ?

Assume equal probabilities for boys and girls ?

Question 3
2 Points
The heights of persons with No. of persons is given , Heights (in Cm ): 58, 59, 60, 61, 62, 63, 64, 65 and No. of persons : 10, 18, 30, 42,35, 28, 16, 8 .

Find Karl Pearson’s coefficient of correlation.

Question 4
2 Points
In a sample of 1000 cases, the mean of a certain test is 14 and S.D is 2.5. Assuming the distribution to be normal, find

(1) how many students score between 12 and 15?

(2) How many score above 18?

(3) how many score below 8?

(4) how many score 16?

Question 5
2 Points
In a Bolt factory, machines A, B and C manufacture respectively 25%, 35% and 40% . Of the total of their outputs 5,4 and 2 % are defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured by machines A, B and C ?

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    1 Answer

    2. I'M ADMIN
      This answer was edited.

      Q//1. (i) not more than three will be busy?

      answer:

      (i) not more than three will be busy?

      answer:

      Solve for P(x<=3) = P(x=0) + P(x=1)+ P(x=2)+P(x=3)

      using formula nCr * (p)^r * (q)^n-r

      where n= 6 , p = 1/15 and  q= 14/15

      P(x<=3) = [1-P(x≥3)]+P(x=3)

      P(x<=3) =0.99492+[6C3 * (1/15)^3 * (14/15)^3]

      P(x<=3) =0.99492+0.00481=0.99973

       

       

      (ii) at least three of them will be busy ?

      Answer: P(x≥3) = P(x=3) + P(x=4)+ P(x=5)+P(x=6)

      P(x≥3)= P(x=3) + P(x=4)+ P(x=5)+P(x=6)

      P(x≥3)=[6C3 * (1/15)^3 * (14/15)^3 ]+[6C4 * (1/15)^4 * (14/15)^2 ]+[6C5 * (1/15)^5 * (14/15)^1 ]+[6C6 * (1/15)^6 * (14/15)^0 ]

      P(x≥3)=0.00508

       

       Q2//  answer: go through the below link

      https://sikshapath.in/question/out-of-800-families-with-4-children-each-how-many-families-would-be-expected-to-have-1-two-boys-2/  

       

      Q//3. answer: 

      https://sikshapath.in/question/the-heights-of-persons-with-no-of-persons-is-given-heights-in-cm-58-59-60-61-62/  

       

      Q//4. answer: click on sample case to get pdf —>sample case

       

      Q//5. answer: click on bolt factory to get pdf —> bolt factory  

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