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A study by the department of Environmental Protection of NJ showed that the average number of airplanes flying through the town of Wayne is four per hour. Assume the passing of these airplanes through...

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A study by the department of Environmental Protection of NJ showed that the average number of airplanes flying through the town of Wayne is four per hour. Assume the passing of these airplanes through Wayne is approximated by the Poisson distribution.

(a) Find the probability that no airplanes flew between 3am and 4am on Monday.

(b) Find the probability that exactly three airplanes flew during this period.

(c) Find the probability that eight or more airplanes flew during the same period.

3. A study by Miami University showed that 40% of baseball fans attending a Marlins baseball game will return for future games. Suppose seven fans are selected at random, what is the probability that:

(a) Exactly four fans will return?

(b) All seven fans will return?

(c) At least six fans will return?

(d) At least one fan will return?

(e) How many fans would be expected to return for future games?

4. The Management of the Tinton Falls Outlets conducted a study on the time customers spent shopping. The study showed that the customers spent an average of 1.2 hours shopping, with a standard deviation of 0.40 hours and it follows the normal probability distribution.

(a) Find the portion of the customers who were shopping between 1.2 and 2 hours.

(b) Find the portion of the customers who were shopping more than 2 hours.

(c) Find the portion of the customers who were shopping between 2 and 2.2 hours.

(d) Find the portion of the customers who were shopping between 1 and 2.2 hours.

Answered Same Day Jan 02, 2022

Solution

David answered on Jan 02 2022
105 Votes
transtutors.com/questions/-2303999.htm
in
(Question 2)
A P (X = 0),λ = 4
P (X = 0) = e
−λλk
k! =
e−4(4)0
0! =
(0.018 316) × (1)
1 = 0.018 316
Use excel function POISSON.DIST( X, mean, cumulative) to calculate PX(0):
POISSON.DIST( 0, 4, FALSE ) = 0.018 315 638 888 7
P (X = 0) = 0.0183
B P (X = 3),λ = 4
P (X = 3) = e
−λλk
k! =
e−4(4)3
3! =
(0.018 316) × (64)
6 = 0.195 367
Use excel function POISSON.DIST( X, mean, cumulative) to calculate PX(3):
POISSON.DIST( 3, 4, FALSE ) = 0.195 366 814 813
P (X = 3) = 0.1954
C P (X ≥ 8),λ = 4
P (X ≥ 8) = 1 − P (0 ≤ X ≤ 7) where P (0 ≤ X ≤ 7) =P (X = 0) + P (X = 1)
+ P (X = 2)+... + P (X = 7)
P (X = 0) = e
−λλ0
0! =
e−4(4)0
0! = 0.018 316
P (X = 1) = e
−λλ1
1! =
e−4(4)1
1! = 0.073 263
P (X = 2) = e
−λλ2
2! =
e−4(4)2
2! = 0.146 525
...
P (X = 7) = e
−4(4)7
7! = 0.059 540
Answer = 1 − (0.018 316 + 0.073 263 + 0.146 525 + ... + 0.059 540) = 0.051 134
Use excel function POISSON.DIST( X, mean, cumulative) to calculate
P (X ≥ 8) = 1 − P (X ≤ 7) = 1 − FX(7):
1-POISSON.DIST( 7, 4, TRUE ) = 0.051 133 615 792 8
P (X ≥ 8) = 0.0511
X
(Question 3)
A P (X = 4),n = 7, p = 0.4
X ∼ Binomial(n = 7, p = 0.4)
P (X = 4) =
(
n
x
)
px(1 − p)n−x =
(7
4
)
(0.4)4(1 − 0.4)7−4 = 35 ∗ (0.4)4(0.6)3 =
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transtutors.com/questions/-2303999.htm
in
0.193536 = 0.193536 ≈ 0.194
P (X = 4) = 0.194
Using excel function BinomDist(4,7,0.4,false) or TI-83/84 function
inompdf(7,0.4,4), exact answer is 0.193536
B P (X = 7),n = 7, p = 0.4
X ∼ Binomial(n = 7, p = 0.4)
P (X = 7) =
(
n
x
)
px(1 − p)n−x =
(7
7
)
(0.4)7(1 − 0.4)7−7 = 1 ∗ (0.4)7(0.6)0 =
0.001638 ≈ 0.001638 ≈ 0.002
P (X = 7) = 0.002
Using excel function BinomDist(7,7,0.4,false) or TI-83/84...
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