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9Criterion A Title Is the height of the grass in my backyard normally distributed A staple of suburban neighborhoods is the distant sound of lawnmowers humming and the lingering scent of fresh cut grass I was weeding in my backyard when I spotted a patch of long grass that look as though it hadn't been mowed in quite some time I wondered how tall untrimmed grass could really grow to and if there was a normal distribution among the heights of the blades as it is an uncommon spectacle in most suburban neighborhoods The main purpose of this investigation is to determine whether the heights of the blades of uncut grass in my backyard are normally distributed I will do so by collecting 125 blades of uncut grass from a patch of my backyard and measuring their heights in centimeters Next I will create a table of just the heights of the grass to the nearest millimeter With this data I will create a grouped frequency table displaying their heights in intervals of 1 centimeter Then I will make a histogram graph to look for a bell shaped pattern that would indicate normal distribution After making a histogram I will calculate the mean height and standard deviation of the data I will compare these measurements to those of a normally distributed graph

Then I will calculate the expected values of the frequency table and compare them to the observed values that I have collected Next I will use the chi squared formula with a significance level of 5 comparing the critical value of my data found using significance level and degrees of freedom to my chi squared value If the chi squared calculated value is less than the critical value I will accept the null hypothesis that my data follows a normal distribution Criterion B I collected 125 blades of grass from an untrimmed patch in my backyard I used scissors to cut the blades as close to the ground as I could I then taped each blade flat on a piece of paper and measured their heights in centimeters starting from the bottom where they had been cut up to their tallest point The heights of the 125 blades of grass are as follows measured in centimeters to the nearest millimeter I then grouped my data into an interval frequency table and using

Excel created a histogram to represent the heights of the blades of grass to find out if they followed a normal distribution Just by looking at the graph it appears to be positively skewed there is no clear bell shape to indicate a normal distribution Criterion C To mathematically check for a normal distribution I will start by finding the mean and standard deviation of my data The total height of the blades of grass is 3213 cm Therefore the mean height of the grass is 3213 125 25 704 cm The standard deviation is 4 853 cm In a normal distribution bell shaped graph 68 26 of the data should be between one standard deviation above and below the mean 95 44 between two standard deviations of the mean and 99 74 between three standard deviations of the mean In my data set 1 standard deviation from the mean is 25 704 4 853 20 851 cm and 25 704 4 853 30 557 cm Between 20 851 and 30 557 cm are 83 blades of grass 83 125 100 66 4 This is somewhat far from the 68 26 that the amount of data that falls into this interval should be 2 standard deviations from the mean is 25 704 2 4 853 15 998 cm and 25 704 2 4 853 35 41 cm Between 15 998 and 35 41 cm are 120 blades of grass 120 125 100 96 This is very near to the 95 44 that the amount of data that falls into this interval should be 3 standard deviations from the mean is 25 704 3 4 853 11 145 cm and 25 704 3 4 853 40 263 cm Between 11 145 and 40 263 cm are all 125 blades of grass or 100 of the data

This is again very near to the 99 74 that the amount of data that falls into this interval should be Using the chi squared test of independence χ 2 at a significance level of 5 I will determine if my data follows a normal distribution To find the expected values of my data I will use the use the normCDF function of my TI nspire for each height interval of the frequency table entering its lower and upper bounds the mean height of the total data 25 704 and the standard deviation 4 853 This function will give me a decimal representing the probability of data that falls within the selected height interval I will multiply this given decimal by 125 total blades collected giving me the expected value which I will round to the nearest tenth In the interval for example I will enter 20 for the lower bound 21 for the upper bound 25 704 for the mean and 4 853 for the standard deviation The format of this in the TI nspire is normCDF 20 21 25 704 4 853 0 046272 0 046272 125 an expected value of 5 78 for the interval the data follows a normal distribution with mean 25 704 cm and standard deviation 4 853 cm the data does not follow a normal distribution with mean 25 704 cm and standard deviation 4 853 cm Because I calculated the mean and standard deviation of my data the degrees of freedom is n 3 instead of n 1 So there are 11 degrees of freedom The critical value of 11 degrees of freedom with a significance level of 5 is 19 675 The chi squared formula is 13 1733 is less than 19 675 so I accept the null hypothesis

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