A survey asked high school cell phone users how many minutes they use their phones to talk each day. The
1 answer:
The two numbers in between which the lower quartile value given by the attached boxplot falls are 1.5 and 2.5 respectively.
- The lower quartile means the 25th percentile or lower
of the distribution
- From a boxplot, the lower quartile is the point marked by the starting point of the box.
- The Lower quartile of the distribution represented by the boxplot is 2
- The values which bounds the lower quartile value to the left and right are 1.5 and 2.5 respectively.
Therefore, two numbers in between which the lower quartile value falls are 1.5 and 2.5
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In order to make a table, we sample some x values (whichever we want), and we compute the expression for those value. Each x value will yield a unique y value.
If you need this table to graph the function, you'll only need two points, since this is a line, and having two points you just need to connect them.
Here are some samples, feel free to make more if you need to:
So, we have the following table
Answer:
(a) P (X = 0) = 0.0498.
(b) P (X > 5) = 0.084.
(c) P (X = 3) = 0.09.
(d) P (X ≤ 1) = 0.5578
Step-by-step explanation:
Let <em>X</em> = number of telephone calls.
The average number of calls per minute is, <em>λ</em> = 3.0.
The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.
The probability mass function of a Poisson distribution is:
(a)
Compute the probability of <em>X</em> = 0 as follows:
Thus, the probability that there will be no calls during a one-minute interval is 0.0498.
(b)
If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.
Compute the probability of <em>X</em> > 5 as follows:
P (X > 5) = 1 - P (X ≤ 5)
Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.
(c)
The average number of calls in two minutes is, 2 × 3 = 6.
Compute the value of <em>X</em> = 3 as follows:
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Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.
(d)
The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.
Compute the probability of <em>X</em> ≤ 1 as follows:
P (X ≤ 1 ) = P (X = 0) + P (X = 1)
Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.