PLEASE HELP ASAP THIS IS A MAJOR TEST FOR ME! Below is the number of innings pitched by each of the Greenbury Goblins' six start
ing pitchers during the Chin Tournament. Find the median number of innings pitched.
1 answer:
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Answer:
(a) The value of P (None) is 0.062.
(b) The value of P(at least one) is 0.938.
(c) The value of P(at most one) is 0.253.
(d) The event is not unusual.
Step-by-step explanation:
Let <em>X</em> = number of households watching the show.
The probability of the random variable <em>x</em> is, P (X) = <em>p</em> = 0.18.
The sample selected for the survey is of size, <em>n</em> = 14
The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 14 and <em>p</em> = 0.18.
The probability of a Binomial distribution is computed using the formula:
(a)
Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:
Thus, the value of P (None) is 0.062.
(b)
Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:
P (X ≥ 1) = 1 - P (X < 1)
= 1 - P (X = 0)
Thus, the value of P(at least one) is 0.938.
(c)
Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:
P (X ≤ 1) = P (X = 0) + P (X = 1)
Thus, the value of P(at most one) is 0.253.
(d)
An event that has a very low probability of occurrence is known as an unusual event.
The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.
This probability value is not low.
Hence, the event is not unusual.