Jawaban Soal Statistika CH 6

15 Maret 2023 Jawaban Soal Statistika CH 6

NAMA: Ahmad Fahre Zamil Saputra NIM : 23080694409

KELAS : 2023 B

REZA FAHREZA

UNIVERSITAS NEGERI SURABAYA

.194

3.Compute the mean and variance of the following probability distribution.

JAWAB:

X P(x) X.P(x) [(x-µ)²P(x)]

5 .1 0.5 (5-14,5)²(.1) = 9,025

10 .3 3 (10-14,5)²(.3) = 6,075

15 .2 3 (15-14,5)²(.2) = 0,05

20 .4 8 (20-14,5)²(.4) = 12,1

Total 14,5

Mean = 14,5 Found by:

𝜇 = 5(. 1) + 10(. 3) + 15(. 2) + 20(. 4) = 14,5

Variance = 27,25 Found by:

𝜎² = ∑[(𝑥 − 𝜇)²𝑃(𝑥)]

𝜎² = 9,025 + 6.075 + 0,05 + 12,1 = 27,25

4.Which of these variables are discrete and which are continuous random variables?

a.The number of new accounts established by a salesperson in a year.

ANSWER: Discrete random variable

b. The time between customer arrivals to a bank ATM.

ANSWER: Continuous random variable

c. The number of customers in Big Nick’s barber shop.

ANSWER: Discrete random variable d.The amount of fuel in your car’s gas tank.

ANSWER: Continuous random variable e. The number of minorities on a jury.

ANSWER: Discrete random variable f. The outside temperature today.

ANSWER: Continuous random variable

5. The information below is the number of daily emergency service calls made by the volunteer ambulance service of Walterboro, South Carolina, for the last 50 days. To explain, there were 22 days on which there were 2 emergency calls, and 9 days on which there were 3 emergency calls.

a.Convert this information on the number of calls to a probability distribution.

ANSWER:

X Frequency P(X) X.P(X) [(x-µ)²P(x)]

0 8 0.16 0 0.4624

1 10 0.20 0.20 0.0980

2 22 0.44 0.88 0.0396

3 9 0.18 0.54 0.3042

4 1 0.02 0.08 0.1058

Total 50 1.70 1.0100

b. Is this an example of a discrete or continuous probability distribution?

ANSWER:

Distribusi diskrit, karena variable X hanya mengambil hasil yang berbeda.

c.What is the mean number of emergency calls per day?

ANSWER:

𝜇= Σ𝑥.𝑃(𝑥)

= 0.(0,16) + 1.(0.2) + 2.(0,44) + 3.(0,18) + 4.(0,02)

= 0 + 0,2 + 0,88 + 0,54 + 0,08

= 1,70

d.What is the standard deviation of the number of calls made daily?

ANSWER:

𝐸[𝑥2] = Σ[𝑥2𝑃(𝑥)]

= (0²)(0,16) + (1²)(0,2) + (2²)(0,44) + (3²)(0,18) + (4²)(0,02)

= 0 + 0,2 + 1,76 + 1,62 + 0,32

= 3,9

Variance = 𝐸[𝑥2]−(𝐸[𝑥])2

= 3,9 – (1,7)²

= 3,9 – 2,89

= 1,01

Standart deviation = √𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒

= √1,01

= √1,005

p. 200

10.In a binomial situation, n 5 and .40. Determine the probabilities of the following events using the binomial formula.

a. x 1 ANSWER:

𝑃(𝑥 = 1) = 5∁1 × 0,40¹× (1 − 0,40)⁽⁵⁻¹⁾ = 5 × 0,40 × 0,1296 = 0,2592 Probability = 0.2592

b. x 2 ANSWER:

𝑃(𝑥 = 1) = 5∁2 × 0,40²× (1 − 0,40)⁽⁵⁻²⁾

= 10 × 0,16 × 0,216 = 0,3456

Probability = 0,3456

12.Assume a binomial distribution where n = 5 and π=0,30.

a.Refer to Appendix B.9, and list the probabilities for values of x from 0 to 5.

ANSWER:

X Formula P(X)

0 5∁0 × 0,30 × 0,75 0,168

1 5∁1 × 0,30 × 0,74 0,360

2 5∁2 × 0,30 × 0,73 0,309

3 5∁3 × 0,30 × 0,72 0,132

4 5∁4 × 0,30 × 0,71 0,028

5 5∁5 × 0,30 × 0,70 0,002

b.Determine the mean and standard deviation of the distribution from the general definitions given in formulas (6–1) and (6–2).

ANSWER:

𝜇 = 0 × 0,168 + 1 × 0,360 + 2 × 0,309 + 3 × 0,132 + 4 × 0,028 + 5 × 0,002 𝝁 = 𝟏, 𝟒𝟗𝟔

𝜎 = √1,0417 𝝈 = 𝟏, 𝟎𝟐𝟎𝟔