We read your photo as:

Abe plays for his school's cricket team. The number of runs scored by Abe in the eight games that he batted in, is shown below. (Abe was given out in all of the games.) 21 8 19 7 15 32 14 12 1.1.1 Determine the variance of the data. 1.1.2 Calculate the standard deviation 1.1.3 Interpret what the standard deviation tells us about the spread of the data. 1.2 The following table shows the number of hours studied and corresponding test scores of five students: Hours Studied (x) Test Scores (y) 2 50 4 65 6 70 8 85 10 90 1.2.1 Plot a scatter plot in ADDENDUM A using the given data 1.2.2 Describe the type of correlation shown in the scatter plot. 1.2.3 Draw an intuitive line of best fit on your graph and explain its significance. 1.2.4 Find the equation of the regression line in the form y = bx + a. Use one of the two tables on ADDENDUM B (attached) and then complete all the calculations. 1.2.5 Predict the test score for a student who studies for 7 hours using your equation. 1.2.6 Explain how regression analysis helps in making predictions.

Mathematics

ScanSolve AI Answer

Answers

3 parts
1.1

57116\dfrac{571}{16}

1.5

y=13019x+99519y = \dfrac{130}{19}x + \dfrac{995}{19}

1.6
73

Step-by-step solution

9 steps

Part 1.1· 3 steps

Determine the variance of the data (runs: 12, 14, 15, 18, 19, 20, 24, 32)

  1. Calculate ∑xi\sum x_i and ∑xi2\sum x_i^2.

    ∑xi=12+14+15+18+19+20+24+32=154\sum x_i = 12 + 14 + 15 + 18 + 19 + 20 + 24 + 32 = 154 ∑xi2=122+142+152+182+192+202+242+322=144+196+225+324+361+400+576+1024=3250\sum x_i^2 = 12^2 + 14^2 + 15^2 + 18^2 + 19^2 + 20^2 + 24^2 + 32^2 = 144 + 196 + 225 + 324 + 361 + 400 + 576 + 1024 = 3250

  2. Calculate the mean xˉ\bar{x}.

    xˉ=∑xin=1548=774\bar{x} = \frac{\sum x_i}{n} = \frac{154}{8} = \frac{77}{4}

  3. Calculate the population variance σ2\sigma^2.

    σ2=∑xi2n−xˉ2=32508−(774)2=650016−592916=57116\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{3250}{8} - \left( \frac{77}{4} \right)^2 = \frac{6500}{16} - \frac{5929}{16} = \frac{571}{16}

    \dfrac{571}{16}

Part 1.2· 1 step

Interpret the standard deviation

  1. Calculate the standard deviation σ\sigma.

    σ=σ2=57116=5714≈5.97\sigma = \sqrt{\sigma^2} = \sqrt{\dfrac{571}{16}} = \dfrac{\sqrt{571}}{4} \approx 5.97

    The standard deviation of approximately 5.97 runs shows the typical deviation of runs scored from the mean of 19.25 runs, measuring the spread or consistency of A's performance across the 8 games.

    Data for hours studied (xx) and test scores (yy): (0, 50), (2, 70), (5, 85)

Part 1.3

Plot the scatter diagram

The scatter plot has points at $(0, 50)$, $(2, 70)$, and $(5, 85)$, showing an upward trend.

Part 1.4· 1 step

Describe the type of correlation

  1. Observe the pattern.

    As hours studied increase from 0 to 5, test scores increase from 50 to 85.

    There is a strong positive linear correlation between hours studied and test scores.

Part 1.5

y=13019x+99519y = \dfrac{130}{19}x + \dfrac{995}{19}

Find the equation of the best fit line y=mx+cy = mx + c

  1. Calculate the sums.

    n=3,∑x=0+2+5=7,∑y=50+70+85=205n = 3, \quad \sum x = 0 + 2 + 5 = 7, \quad \sum y = 50 + 70 + 85 = 205 ∑x2=02+22+52=0+4+25=29\sum x^2 = 0^2 + 2^2 + 5^2 = 0 + 4 + 25 = 29 ∑xy=0⋅50+2⋅70+5⋅85=0+140+425=565\sum xy = 0 \cdot 50 + 2 \cdot 70 + 5 \cdot 85 = 0 + 140 + 425 = 565

  2. Calculate slope mm.

    m=n∑xy−(∑x)(∑y)n∑x2−(∑x)2=3⋅565−7⋅2053⋅29−72=1695−143587−49=26038=13019m = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum x^2 - (\sum x)^2} = \frac{3 \cdot 565 - 7 \cdot 205}{3 \cdot 29 - 7^2} = \frac{1695 - 1435}{87 - 49} = \frac{260}{38} = \frac{130}{19}

  3. Calculate intercept cc.

    c=∑y−m∑xn=205−13019⋅73=205−910193=3895−910193=2985193=298557=99519c = \frac{\sum y - m \sum x}{n} = \frac{205 - \frac{130}{19} \cdot 7}{3} = \frac{205 - \frac{910}{19}}{3} = \frac{\frac{3895 - 910}{19}}{3} = \frac{\frac{2985}{19}}{3} = \frac{2985}{57} = \frac{995}{19}

    The equation is y=13019x+99519y = \frac{130}{19} x + \frac{995}{19}.

    y = \dfrac{130}{19}x + \dfrac{995}{19}

Part 1.673

Predict the test score for B who studied 3 hours

  1. Substitute x=3x = 3 into the regression equation.

    y=13019⋅3+99519=390+99519=138519≈72.89y = \frac{130}{19} \cdot 3 + \frac{995}{19} = \frac{390 + 995}{19} = \frac{1385}{19} \approx 72.89

    Predicted test score is approximately 73 marks.

    73

Part 1.7

Explain how regression analysis makes predictions

Regression analysis finds the best fit line y=mx+cy = mx + c by minimizing the sum of squared differences between observed yy values and predicted y^\hat{y} values on the line. Predictions are made by substituting a given xx value into this equation to find y^\hat{y}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QAbe plays for his school's cricket team. The number of runs scored by Abe in the eight games that he…QThe table shows hours studied and test scores for five students. Plot a scatter plot, describe the…QThe following is record of scores obtained 30 J.H.S. 2 pupils in a test marked out of 5: 5, 4, 3, 1…QThe table shows marks scored by students: a) Calculate the mean mark. b) The table below shows time…QA teacher is analysing the study habits of students. The number of hours that students studied per…
Browse by topic:Statistics
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.