ACTIVITY 2 by GIAN CLEESE BOGUIREN

giancleese_boguiren_pharma — 2018-09-19 08:32:25 UTC

1. Suppose that in a bowling tournament, the scores among all bowlers are normally distributed with mean = 182 with a standard deviation = 14.

giancleese_boguiren_pharma — 2018-09-19 08:33:16 UTC

A. What proportions of the players scored less than 175 points?

Given:

μ = 182

σ = 14

x = 175

Required:

(P<175)

Solution:

(P<175)

Z= (175-182) / 14

Z= -0.5

-0.5 = 0.30854

0.30854 x 100 = 30.85%

Answer:

There are 30.85% proportions of the players scored less than 175 points.

giancleese_boguiren_pharma — 2018-09-19 08:34:24 UTC

giancleese_boguiren_pharma — 2018-09-19 08:36:55 UTC

B. What proportions of the players scored more than 200 points?

Given:
μ = 182
σ = 14
x = 200

Required:
(P>200)

Solution:
(P>200)

Z= (200-182) / 14
Z= 1.29
1.29= 0.90147
1-0.90147= 0.09853
0.09853 (100)= 9.85%

Answer:
There are 9.85% proportions of the players scored more than 200 points.

giancleese_boguiren_pharma — 2018-09-19 08:45:42 UTC

C. What proportions of the players scored between 180 and 210 points?

(1)

Given:

μ = 182

σ = 14

Required:

Proportions of the players that scored between 180 and 210

Solution:

(180)

x = z - μ /σ

x = (180 -182) /14

x = -0.14

(Area under normal curve)

-0.14 = .44433

.44433 (100) = 44.433%

(2)

Given:

μ = 182

σ = 14

Required:

Proportions of the players that scored between 180 and 210

Solution:

(210)

x = z - μ /σ

x = (210 -182) /14

x = 2

(Area under normal curve)

2 = .97725

.97725 (100) = 97.725%

97.725% - 44.433% = 53.29%

Answer:

There are 53.29% proportions of the players that scored between 180 and 210 points.

giancleese_boguiren_pharma — 2018-09-19 08:47:47 UTC

giancleese_boguiren_pharma — 2018-09-19 08:59:02 UTC

2. The running time for videos submitted on YouTube in a given week is normally distributed with mean, = 390 seconds with a standard deviation, = 148 seconds.

giancleese_boguiren_pharma — 2018-09-19 08:59:55 UTC

A. If a single video is selected at random, what is the probability that the running time of the video exceeds 6 minutes?

Given:

μ = 390 seconds

σ = 148 seconds
x = 6 minutes = 360 seconds

Required:

The probability that the running time of the videos exceed 6 minutes

Solution:

Z= (360-390) / 148

Z= - 0.2027 = 0.42074

0.5 - 0.42074 = 0.07926

0.5 + 0.07926 = 0.57926 = 57.926%

Answer:

The probability of selecting a video that exceeds 6 minutes is 57.93%.

giancleese_boguiren_pharma — 2018-09-19 09:03:54 UTC

B. If a single video is selected at random, what is the probability that the running time of the video is less than 6.2 minutes?

Given:
μ = 390 sec
σ =148 sec

Required:
probability that the running time of the video is less than 6.2 minutes.

Solution:

Z= (372-390) / 148

Z= -0.1216

(area under normal curve)

0.0478

50 - 4.78

Answer:

The probability of selecting a video that is less than 6.2 minutes is 45.22%

giancleese_boguiren_pharma — 2018-09-19 09:07:29 UTC

C. If a single video is selected at random, what is the probability that the running time of the video is between 6.8 minutes and 10.2 minutes?

Given:

μ = 390 sec

σ = 148 sec

X1 = 6.8 min = 408 sec

X2 = 10.2 minutes = 612 sec

Required:

P(6.8min

Solution:

Z1= x1-= (408-390) / 148 =0.12

= 0.54776 x 100 = 54.776%

Z2 = x2-= (612-390) / 148 = 1.5 = 0.93319 x 100 = 93.319%

Zt = Z2 - Z1 = 93.319 - 54.776 = 38.543%

Answer:

The probability of selecting a video that is between 6.8 minutes and 10.2 minutes is 38.54%.

SUBMITTED BY: GIAN CLEESE A. BOGUIREN (9)

giancleese_boguiren_pharma — 2018-09-19 09:16:23 UTC