# Khan/khan-exercises

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 Z-scores 2: Reading a z-table
randRange(0, 9) randRange(0, 9) roundTo(2, randRange(0, 2) + ROW_INDEX / 10 + COL_INDEX / 100) randRange(70, 90) randRange(2, 6) roundTo(2, GRADE - ZSCORE * STDDEV) (function() { var rowNames = []; for(var i = floorTo(0, ZSCORE); i < (floorTo(0, ZSCORE) + 1); i += 0.1) { rowNames.push(i.toFixed(1)); } return rowNames; })() [".00", ".01", ".02", ".03", ".04", ".05", ".06", ".07", ".08", ".09"] (function() { var zGrid = []; for (var i = 0; i < ROWS.length; i++) { var zRow = []; for (var j = 0; j < COLUMNS.length; j++) { zRow.push(zScores(roundTo(2, (floorTo(0, ZSCORE) + i / 10 + j / 100) * 100)).toFixed(4)); } zGrid.push(zRow); } return zGrid; })() ZGRID[ROW_INDEX][COL_INDEX] "z"
The scores on a statewide course(1) exam were normally distributed with \mu = MEAN and \sigma = STDDEV.
person(1) earned an GRADE on the exam.

person(1)'s exam grade was higher than what percentage of test-takers? Use the cumulative z-table provided below. Round to two decimal places.

rowzgrid
roundTo(2, ANSWER * 100)

A cumulative z-table shows the probability that a standard normal variable will be less than a certain value (z).

In order to use the z-table, we first need to determine the z-score of person( 1 )'s exam grade.

Recall that we can calculate his( 1 ) z-score by subtracting the mean (\mu) from his(1) grade and then dividing by the standard deviation (\sigma).

Look up ZSCORE on the z-table. This value, ANSWER, represents the portion of the population that scored lower than GRADE on the exam.

var nth = ":nth-child(" + (COL_INDEX + 2) + ")"; $(".fake_row").eq(ROW_INDEX).find("span").css("background", ORANGE);$(".fake_row span" + nth).css("background", ORANGE); $(".fake_header span" + nth).css("background", ORANGE);$(".fake_row").eq(ROW_INDEX).find("span" + nth).css("background", BLUE);

person( 1 ) scored higher than roundTo(2, ANSWER * 100)\% of the test-takers on the course( 1 ) exam.

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