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Submitted by: Joslean Grace Galera    1RMT

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Submitted by: Joslean Grace Galera    1RMT

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

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

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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?

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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?

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Given:

μ= 390 sec

σ=148 sec

X1 = 6.8 min = 408 sec

X2 = 10.2 minutes =612 sec

Required:

P(6.8min<Z<10.2min)

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%.

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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? 

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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? 

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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 = 45.22

Answer:

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

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

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

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Given:
μ(mean) = 390 seconds

σ(sd)  = 148 seconds

x  =  6 minutes = 360 seconds

Required:

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

Solution:

Z= (x - μ) / σ

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%.

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

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

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PART 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%

PART 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:

53.29%

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

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

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Given:
μ(mean) = 182

σ(sd)  = 14

x  =  175
Required:

proportions of the player scored more than 200 points

Solution:

(P>200)

Z= x- μ / σ

Z=  200-182 / 14

Z= 1.29 (Area under normal curve)

1.29= 0.90147

1.00 - 0.90147= 0.09853

0.09853 (100)= 9.853%

Answer:

9.85%

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

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

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Given:

μ(mean) = 182

σ(sd)  = 14

x  =  175

Required:  proportions of the players scored less than 175 points
Solution:

(P<175)

Z= x- μ / σ

Z=  175-182 / 14

Z= -0.5 (Area under normal curve)

-0.5 = 0.30854

0.30854 x 100 = 30.85%

Answer:

30.85%

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

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