Please help, I need this by 10 30 AM
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Answer:
There are 6 ways to run 10 m between two cones of the four cones giving a maximum available distance to run as 60 meters
Step-by-step explanation:
The parameters given are
Distance between two cones = 10 m
Number of cones = 4
Therefore, number of ways to choose two 10 m distances (distance between two cones) from the four cones = = 6
Hence there are only six 10 m distances to run between two cones which gives a maximum distance available to run as 60 m.
Answer:
(a) The value of P (X ≤ 2) is 0.8729.
(b) The value of P (X ≥ 5) is 0.0072.
(c) The value of P (1 ≤ X ≤ 4) is 0.7154.
(d) The probability that none of the 25 boards is defective is 0.2774.
(e) The expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.
Step-by-step explanation:
The random variable <em>X</em> is defined as the number of defective boards.
The probability that a circuit board is defective is, <em>p</em> = 0.05.
The sample of boards selected is of size, <em>n</em> = 25.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The probability mass function of <em>X</em> is:
(a)
Compute the value of P (X ≤ 2) as follows:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
Thus, the value of P (X ≤ 2) is 0.8729.
(b)
Compute the value of P (X ≥ 5) as follows:
P (X ≥ 5) = 1 - P (X < 5)
Thus, the value of P (X ≥ 5) is 0.0072.
(c)
Compute the value of P (1 ≤ X ≤ 4) as follows:
P (1 ≤ X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
Thus, the value of P (1 ≤ X ≤ 4) is 0.7154.
(d)
Compute the value of P (X = 0) as follows:
Thus, the probability that none of the 25 boards is defective is 0.2774.
(e)
Compute the expected value of <em>X</em> as follows:
Compute the standard deviation of <em>X</em> as follows:
Thus, the expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.