Stat1250_SGTA5

1 Stat1250 SGTA 5: The Normal Distribution © Copyright Macquarie University In this SGTA we will work with Normal distribution  to calculate probabilities or  y values for a given probability. The Normal distribution is the most important distribution for continuous numerical variables. It is used as a model for many measurements that occur in nature and its use is further extended as a result of the Central Limit Theorem, which is taught in lectures in coming weeks. The Normal distribution has a symmetric “bell” shape. It is described by two parameters; the centre of the distribution, μ, and its standard deviation, σ, a measure of spread. In Lecture 4 you were shown that, for a numerical variable which is known to have a Normal distribution where the mean and standard deviation are also known, you can find areas (i.e. probabilities) under any Normal curve, above or below any value y. That is, you can find the probability that a randomly selected value lies above or below any value y, or between any 2 values y 1 and y 2. We will do that by using statistical software. In STAT1250 that software is Excel. There are an infinite number of Normal distributions. Each one has the same “bell” shape with the values of μ, the population mean, and σ, the population standard deviation, which vary. Recall that in Lecture 4 we wanted to find the probability that the weight of a box of cereal is greater than 451g if the cereal box weights are normally distributed where mean (μ =) 450.5g and standard deviation ( σ =) 0.2g. (note that the diagram is not to the scale, the shaded area is much smaller than it is shown below). 450.5 451 y The area to the right of y = 451 can be calculated in Excel using the NORM.DIST command. NORM.DIST gives the area on the left of the given y which will be the unshaded area for the above diagram. We will use properties of normal distribution to obtain the probability for the shaded area.  The total area under the whole curve = 1  The Normal distribution is symmetric, so the area above z is equal to the area below - z Prob(y < 450) = Prob(y > 451) = NORM. DIST(451, 450.5, 0.2, TRUE) = 1 – NORM. DIST(451, 450.5, 0.2, TRUE) = 0.0062 We can also use Excel to obtain the value of y for a given probability (i.e. we know the probability but we do not know what the corresponding y value is). For such calculations, we will use NORM.INV command. For example, the weight of a cereal box which is at the 90 th percentile of cereal boxes for this population can be calculated in Excel as following y = NORM. INV(0.9, 450.5, 0.2) = 450.76g A box of cereal weighing 450.76g is heavier than 90% of the cereal boxes in this population. You are expected to read this material and think about the problems before coming to class. Participation in SGTAs is a hurdle requirement, you need to participate in at least 10 (see unit guide). SGTA 5: The Normal Distribution

The United Nations Sustainable Development Goals (SDGs) Goal 14: Life Below Water 2 Stat1250 SGTA 5: The Normal Distribution © Copyright Macquarie University  The following steps are a useful for calculating the probability for a given y: 1. Draw a diagram, mark the mean and y value. 2. Shade the required area. 3. Find relevant probabilities using Excel (i.e. use NORM.DIST function/command). 4. Find the required probability (i.e. probability for the shaded area). 5. Write a sentence which summarises your findings.  The following steps are useful for obtaining the y value for a given probability: 1. Draw a diagram, mark the mean. 2. Shade the area for the given probability and mark where you expect y value to be. 3. Use Excel NORM.INV function/command to obtain the required y value. 4. Write a sentence which summarises your findings. In 2018, The Food and Agriculture Organization of the United Nations issued its report: The State of World Fisheries and Aquaculture, http://www.fao.org/3/i9540en/I9540EN.pdf . This report is in support of the sustainable development goals target (14.4). The report states: “ The state of marine fishery resources, based on FAO’s monitoring of assessed marine fish stocks, has continued to decline. The fraction of marine fish stocks fished within biologically sustainable levels has exhibited a decreasing trend, from 90.0 percent in 1974 to 66.9 percent in 2015 .” Effective management of global fisheries and aquaculture are essential to the future viability of fish species and the continued availability of fish as a resource. Management of fish species includes ensuring that only fish of legal length are caught. Larger fish must also be left in the population as larger female fish lay many more eggs than smaller female fish. Let assume that we have found that the population of fish is approximately normally distributed with an average length of 34cm , and a standard deviation of 11cm . If you have access to a computer, use the Excel relevant function (see above) to answer the questions on the next pages. n Probability Code Probability NORM.DIST(34,45,11,TRUE) 0.1587 NORM.DIST(34,28,11,TRUE) 0.7073 NORM.DIST(45,34,11,TRUE) 0.8413 NORM.DIST(95,34,11,TRUE) 1.0000 NORM.DIST(45,34,11,FALSE) 0.0220 NORM.DIST(52,34,11,TRUE) 0.9491 NORM.DIST(34,25,11,TRUE) 0.7934 NORM.DIST(2,34,11 , TRUE) 0.0018 NORM.DIST(25,34,11,TRUE) 0.2066 NORM.DIST(2,0,1,TRUE) 0.97725 NORM.DIST(20,34,11,TRUE) 0.1016 NORM.INV(0.05,34,11) 15.9066 NORM.DIST(28,34,11,TRUE) 0.2927 NORM.INV(0.95,34,11) 52.0934 Solving probability problems

4 Stat1250 SGTA 5: The Normal Distribution © Copyright Macquarie University 4. Any fish whose length is in the top 5% need to be released. What is the length of a fish for which only 5% of fish will be greater than that length? 52.0934 5. If 10 fish are caught from this population, how many would be expected to be greater than 45 cm long? 6. For any variable, y, in a Normal distribution, we can calculate a z score. The z score is a measure of the number of standard deviations that a variable is from its mean. The z score converts the random variable, y, to a standardised random variable, z, which has a mean of 0 and a standard deviation of 1. A z score is calculated using the formula: 𝒛𝒛 = 𝒚𝒚−𝝁𝝁 𝝈𝝈 What is the z score for a fish whose length is 56cm? Using this z score, what is the probability that a fish is more than 56cm in length?