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Statistics and Probability Q
Francis Davis M. Gamotin
CDO-SHS
STEM 11 C
Lesson 1
Ma’am Rufe A. Felicilda Normal
Distribution
What I Know
1. B. 1
- A. Mean
- C. 0.
- B. 0.
- D. 2.
- C. 0. P(-1 < z < 3) = 0.3413 + 0.4987 =
- C. 0. P(-1.5 < z < -1.3) = 0.4332 – 0. = 0.
- D. z = -2.5 and z = 1 P(-2.5 < z < 1) = 0.4987 + 0.3413 =
- B. 0. P(1 < z < 3) = 0.4938 – 0.3413 =
10.C. 0. P(z > 1) = 0.5000 – 0.3413 = 0. 11.D. 0. P(z > -2.5) = 0.5000 + 0.4938 =
12.B. -
Given: μ = 180, σ = 15, X = 150 z = (X - μ)/σ z = (150 - 180)/ z = - 13.A. 1. Given: μ = 180, σ = 15, X = 200 z = (X - μ)/σ z = (200 - 180)/ z = 1. 14.C. 0. 70 th^ = 70% = 0.7000 – 0.5000 =
0.1985 < 0.2000 < 0. where P(z < 0.52) = 0.1985 and P(z < 0.53) = 0. ∣0.2000 – 0.1985∣ = 0.0015, ∣0.2000 – 0.2019∣ = 0. The percentile of the z-value of 0.52 is the nearest to the value of 0.0.2000. z = 0. 15.A. 0. 82 nd^ = 82% = 0.8200 – 0.5000 =
0.3186 < 0.3200 < 0. where P(z < 0.91) = 0.3186 and P(z < 0.92) = 0. ∣0.3200 – 0.3186∣ = 0.0014, ∣0.3200 – 0.3212∣ = 0. The percentile of the z-value of 0.92 is the nearest to the value of 0.0.3200. z = 0.
What’s New: Activity 1
- Given: n = 28, HV = 101, LV = 13 Range = HV – LV R = 101 – 13 R = 88 2 k^ ≥ n 2 k^ ≥ 28 25 ≥ 28 k = 5 classes Class Width = R/k CW = 88/ CW = 17. CW ≈ 18 Test Points in Math Frequency Probability 13 – 30 1 0.0357 or 3.57% 31 – 48 7 0.25 or 25% 49 – 66 11 0.3929 or 39.29% 67 – 84 7 0.25 or 25% 85 – 102 2 0.0714 or 7.14% TOTAL: 28 1
- By looking at the histogram itself, you can tell that the distribution is approximately normal. The graph is bell-shaped and symmetric about the mean, which means we can assume that a normal curve can be drawn over it.
What’s More
1.1 Mean score: 100 1.2 Median score: 100 1.3 Modal score: 100
- Standard Deviation: 2σ = 130
Statistics and Probability Q
Francis Davis M. Gamotin
CDO-SHS
STEM 11 C
Lesson 2
Ma’am Rufe A. Felicilda Areas Under the
Normal Curve
What’s More: Activity 1
- z = 0. The area at the intersection is 0.2357.
- z = 1. The area at the intersection is 0.4066.
- z = 2. The area at the intersection is 0.4812.
- z = 1. The area at the intersection is 0.4554.
What I Can Do
COLUMN A COLUMN B
- z = 0.04 L. 0.
- z = 1.06 V. 0.
- z = 2.8 M. 0.
- z = 2.09 T. 0.
- z = 0.49 C. 0.
- z = 3.02 S. 0.
- z = -0.03 I. 0.
- z = -1.05 A. 0.
- z = -2.22 E. 0. 10.z = -3.78 O. 0. 11.z = -0.13 H. 0. Decoded Message: I L O V E M A T H E M A T I C S 1 2 3 4 5 6 7 8 9 5 6 7 8 1 10 11
Statistics and Probability Q
Francis Davis M. Gamotin
CDO-SHS
STEM 11 C
Lesson 3
Ma’am Rufe A. Felicilda Shaded Region Under the
Normal Curve
What’s More? Activity 1: Challenge Me!
- P(z > -1.5) = 0.5000 + 0.4332 =
- P(z = -1) = 0
Statistics and Probability Q
Francis Davis M. Gamotin
CDO-SHS
STEM 11 C
Lesson 4
Ma’am Rufe A. Felicilda Understanding
the Z- Scores
What’s More: Activity 1
Mathematics: Given: μ 1 = 85, σ 1 = 7, X 1 = 90 Solution: z 1 = (X 1 - μ 1 )/σ 1 z 1 = (90 - 85)/ z 1 = 0. Therefore, the z-value that corresponds to the raw score of 90 in Mathematics is 0.71. Science: Given: μ 2 = 89, σ 2 = 13, X 2 = 94 Solution: z 2 = (X 2 - μ 2 )/σ 2 z 2 = (94 - 89)/ z 2 = 0. Therefore, the z-value that corresponds to the raw score of 94 in Science is 0.38. Conclusion: The z-value that corresponds to the raw score in Mathematics is higher than that in Science, therefore Adrian’s standing in Mathematics is higher than his standing in Science.
What I Can Do: Activity 1
- I’m most likely to submit the A Test, assuming that both tests contain the same amount of total items, as doing so would increase my chances of being accepted for the job. Although the information provided is still not enough for me to convince myself. In order for me to decide with certainty, I would need information about the exact passing mark for each tests and the standing of such raw scores by comparing them relatively.
What I Can Do: Activity 2
Given: μ = 110, σ = 10
- Given: X = 98 Solution: z = (X - μ)/σ z = (98 - 110)/ z = -1.
Statistics and Probability Q
Francis Davis M. Gamotin
CDO-SHS
STEM 11 C
Lesson 5
Ma’am Rufe A. Felicilda Percentiles Under the
Normal Curve
What’s More: Activity 1
- Given: z = -2.3, μ = 100, σ = 10 Find: X and Percentile Solution: Percentile = P(z < -2.3) P(z < -2.3) = 0. 0.4893 = 48% z = (X - μ)/σ -2.3 = (X - 100)/ -23 = X – 100 77 = X Conclusion: The corresponding percentile and x-value are 48% and 77 respectively.
- Given: z = -3.1, μ = 11.5, σ = 1. Find: X and Percentile Solution: Percentile = P(z < -3.1) P(z < -3.1) = 0. 0.4990 = 49% z = (X - μ)/σ -3.1 = (X - 11.5)/1. -3.875 = X – 11. 7.625 = X Conclusion: The corresponding percentile and x-value for the street kids are 49% and 7.625 respectively.
What I Can Do: Activity 1
- 30 th^ = 30% = 0.
where P(z < -0.84) = 0.2995 and P(z < - 0.85) = 0. ∣0.3000 – 0.2995∣ = 0.0005, ∣0.3000 – 0.3023∣ = 0. The percentile of the z-value of -0. is the nearest to the value of 0.3000. z = -0.
- 52 nd^ = 52% = 0.5200 – 0.5000 =
0.0199 < 0.0200 < 0. where P(z < 0.05) = 0.0199 and P(z < 0.06) = 0. ∣0.0200 – 0.0199∣ = 0.0001, ∣0.0200 – 0.0239∣ = 0. The percentile of the z-value of 0. is the nearest to the value of 0.0200. z = 0. - 15 th^ = 15% = 0. 0.1480 < 0.1500 < 0. where P(z < -0.38) = 0.1480 and P(z < - 0.39) = 0. ∣0.1500 – 0.1480∣ = 0.0020, ∣0.1500 – 0.1517∣ = 0. The percentile of the z-value of -0. is the nearest to the value of 0.1500. z = -0.
- 88 th^ = 88% = 0.8800 – 0.5000 =
0.3790 < 0.3800 < 0. where P(z < 1.17) = 0.3790 and P(z < 1.18) = 0. ∣0.3800 – 0.3790∣ = 0.0010, ∣0.3800 – 0.3810∣ = 0. Since both values display the same amount of distance away from 0.3800, then we need to do the interpolation. zn = (1.17 + 1.18)/ z = 1.
The percentile of the z-value of 0.52 is the nearest to the value of 0.0.2000. z = 0. 15.A. 0. 82 nd^ = 82% = 0.8200 – 0.5000 = 0. 0.3186 < 0.3200 < 0. where P(z < 0.91) = 0.3186 and P(z < 0.92) = 0. ∣0.3200 – 0.3186∣ = 0.0014, ∣0.3200 – 0.3212∣ = 0. The percentile of the z-value of 0.92 is the nearest to the value of 0.0.3200. z = 0.
Statistics and Probability Q
Francis Davis M. Gamotin
CDO-SHS
STEM 11 C
Lesson 1
Ma’am Rufe A. Felicilda Random
Sampling
What I Know: Multiple Choice
- B. Lottery Sampling
- C. Sampling Distribution
- B. Sample
- B. 386 Given: N = 11,000, e = 5% or 0. n = N/(1 + Ne^2 ) n = 11,000/[1 + 11,000(0.05^2 )] n = 11,000/28. n = 386
- A. The mean of the sampling distribution of the sample means is equal to 13.2.
Reason: The mean of a population and the mean of the sampling distribution of the sample means are the same.
- D. 220 Given: n = 3, N = 12 NCn = N!/(N – n)!n! 12 C 3 = 12!/(12 – 3)!3! 12 C 3 = 479,001,600/2,177, 12 C 3 = 220
- A. (^) NCn
- C. Normal
- D. 35 10.D. 0. Given: μx̅ = 180, σ = 8, x̅ = 185, n = 15 P(x̅ > 185) z = (x̅ – μx̅)/(σ/√ n ) z = (185 – 180)/(8/√15) z = 5/2. z = 2. P(x̅ > 185) = P(z > 2.42) P(x̅ > 185) = 0.5000 – 0. P(x̅ > 185) = 0.
What’s More: Activity 3. Random Sample or Not?
- Random Sampling
- Random Sampling
- Not Random Sampling
- Not Random Sampling
- Random Sampling
What Is It: Activity 5. Determining the Sample Size
- Given: N = 20,000, e = 0.05; n =? n = N/(1 + N e^2 ) n = 20,000/[1 + 20,000(0.05)^2 ] n = 392.2 or 393
- Solution (Sample Size) Sample Size Category A 1,000 (1,000/20,000) · 393 = 19.65 20 Category B 5,000^ (5,000/20,000) · 393 = 98.25^98 Category C 6,000 (6,000/20,000) · 393 = 117.9 118 Category D 8,000 (8,000/20,000) · 393 = 157.2 157 TOTAL SAMPLE SIZE: 393
Statistics and Probability Q
Francis Davis M. Gamotin
CDO-SHS
STEM 11 C
Lesson 2
Ma’am Rufe A. Felicilda Parameter
and Statistic
What’s New: Activity (Describe Me)
- Parameter
What’s More: Activity 2 (List and Construct)
A.
6 C 3 = 6! / (6 – 3)!3!
6 C 3 = 20 random samples Random Samples Sample Means 9, 12, 15 12 9, 12, 18 13 9, 12, 21 14 9, 12, 24 15 9, 15, 18 14 9, 15, 21 15 9, 15, 24 16 9, 18, 21 16 9, 18, 24 17 9, 21, 24 18 12, 15, 18 15 12, 15, 21 16 12, 15, 24 17 12, 18, 21 17 12, 18, 24 18 12, 21, 24 19 15, 18, 21 18 15, 18, 24 19 15, 21, 24 20 18, 21, 24 21
B.
Sample Means Frequency Probability 12 1 0. 13 1 0. 14 2 0. 15 3 0. 16 3 0. 17 3 0. 18 3 0. 19 2 0. 20 1 0. 21 1 0. TOTAL: 20 1 C.
Statistics and Probability Q
Francis Davis M. Gamotin
CDO-SHS
STEM 11 C
Lesson 4
Ma’am Rufe A. Felicilda Mean and Variance of the Sampling
Distribution of Sample Means
What I Have Learned
Population Sampling Distribution of Sample Means Mean μ = ΣX /N μx̅ = Σ [x̅ · P(x̅)] Variance σ^2 = [Σ(X – μ)^2 ]/N^ σ^2 x̅ = Σ [P(x̅) · (x̅^ – μ)^2 ] Standard Deviation σ = √σ^2 σx̅ = √σ^2 x̅
- The mean of the population and the mean of the sampling distribution of the sample means are pretty much the same result-wise, although they do go through different processes for solving them. In a population, the mean is solved through dividing the sum of all population items by the number of observations in the said population, whereas the mean of the sampling distribution of the sample means is acquired through the sum of all the multiplication of each sample mean and their corresponding probability.
- The sample variance is an estimate of σ^2 , and is very useful in situations where calculating the population variance would be too time consuming. The sample mean is used, which is the only difference in how the sample variance is calculated. The sample variance, unlike the population variance, is simply a sample statistic. It is dependent on the research methodology used and the sample size selected. A new sample or experiment will almost certainly yield a different sample variance, though if your samples are both representative, your sample variances should be good estimates of the population variance and very similar.
- There’s not much difference between them other than that they are acquired through different variance formulas. Both can be solved just by acquiring the positive square root of their respective variances.
- (^) NCn = N! / (N – n)!n! 6 C 3 = 6! / (6 – 3)!3! 6 C 3 = 20 random samples
- Sample Means 18, 22, 25 21. 18, 22, 28 22. 18, 22, 32 24 18, 22, 36 25. 18, 25, 28 23. 18, 25, 32 25 18, 25, 36 26. 18, 28, 32 26 18, 28, 36 27. 18, 32, 36 28. 22, 25, 28 25 22, 25, 32 26. 22, 25, 36 27. 22, 28, 32 27. 22, 28, 36 28. 22, 32, 36 30 25, 28, 32 28. 25, 28, 36 29. 25, 32, 36 31 28, 32, 36 32
Sample Means Frequency Probability P(x̅) 21.67 1 0. 22.67 1 0. 23.67 1 0. 24 1 0. 25 2 0. 25.33 1 0. 26 1 0. 26.33 2 0. 27.33 2 0. 27.67 1 0. 28.33 1 0. 28.67 2 0. 29.67 1 0. 30 1 0. 31 1 0. 32 1 0.
Means x̅ Probabi lity P(x̅) x̅ · P(x̅) 21.67 0.05 1. 22.67 0.05 1. 23.67 0.05 1. 24 0.05 1. 25 0.10 2. 25.33 0.05 1. 26 0.05 1. 26.33 0.10 2. 27.33 0.10 2. 27.67 0.05 1. 28.33 0.05 1. 28.67 0.10 2. 29.67 0.05 1. 30 0.05 1. 31 0.05 1. 32 0.05 1.
TOTAL
1 Σ [x̅ · P(x̅)] = 26. ≈ 27
- σ^2 x̅ = Σ P(x̅) · (x̅ – μ)^2
x̅ P(x̅ ) x̅ - μ (x̅ – μ)^2 P(x̅) · (x̅ – μ)^2
7
-5.33^2
