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KengaRu [80]
3 years ago
5

The stem-and-leaf plot shows kilometers walked by participants in a charity benefit walk. Use it to answer the questions. a. Des

cribe how to find the range of the data set. b. Find the range.
Mathematics
1 answer:
aivan3 [116]3 years ago
4 0

Answer / Explanation

The question is incomplete. It can be found in search engines:

However, kindly find the complete question below:

Question:

The stem and leaf plot shows kilometers walked by participants in a charity benefit walk. Use it to answer the question

12|3 3 6 7 9 9

13| 1 1 4 5 5

14| 0 0 2 3 3 8 8 9

15| 2 2 2 2 2 3 5 5 7

16| 4 5 5 9 9

17|3 5

(a) How many people participated in the walk? Exactly 35 people participated in the walk.

(b)How many of the walkers traveled more than 14 kilometers? About 16-22 people traveled more than 14 kilometers.

Answers:

(a) Exactly 35 people participated in the walk.

(b) About 16-22 people traveled more than 14 kilometers.

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A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome
Ludmilka [50]

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

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Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), <em>p</em> = 0.52

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.

(1)

Compute the probability of overbooking  as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

                                        =P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

=P(X

Thus, the probability that the flight has empty seats is 0.4625.

4 0
3 years ago
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