12:55 94$ Quarter: A Week: A SSLM No. i MELC(s): Illustrates a normal random variable and its characteristics (M11l126P-IlIc-1). identifies regions under the normal curve corresponding to different standard normal values (M11i12SP-IlIc-3). converts a normal random variable to a standard normal variable and vice versa (M111128P-IlIc-4) and Computes probabilities and percentiles using the standard normal table (M11l128P-lllc-d-1}. '9 Objectives: 1. To illustrate normal random variables and their characteristics. 2. To identify the regions under the normal curve. 3. To compute probabilities and percentiles using the standard normal table. > Title of Textme to Study: > Chapter: _ Pages: _ Topic: . Let Us Discover I. NORMAL RANDOM VARIABLE AND ITS CHARACTERISTICS A normal random variable is a set of numerical value ot continuous random variables which is normally distributed. Normal Distribution represents the probability, or the proportion. or the percentage associated with specific sets of measurement values. A normal curve has the following characteristics: The normal curve is bell-shaped. The curve is symmetrical about its center. The mean. median, and mode coincide at the center and are equal to each other. The width of the curve is determined by the standard deviation of the distribution. The tails of the curve flatten out indefinitely along the horizontal axis, always approaching the axis but never touching it. That is. the curve is asymptotic to the baseline. 6. The total area of the normal curve is 1 or 100%. Each hall of the curve is equivalent to an area of 0.5 Or 50%. 9.59.\"? 1 GSC-CtD-LRMSESSLM, v.r. 0.100, Effective June 14,2021 II. IDENTIFYING REGIONS UNDER THE NORMAL CURVE Every region under the normal curve corresponds to the area bounded by pr or z = 0 and the number of units of the z-score is from the center. Each area value is given in percent, which also signies the percentage of date found within the intenral. The area under the curve is always a positive number even if the 2 value is negative. Illusu'ative Example 1: Find the area under the curve below 2 = 2.62 Step 1': Locate the area of the given z-value from the z-table. WWIWW tau-vi: In \"unwound-era run-u w Sfep 2. Compute the area of the required region. P(z 1.17) = 1.000 - 0.8790 = 0.1210 :5090 Step 3: Sketch the normal curve and shade the desired area. Step 4: State the conclusion. :The area above z = 1.17 is 0.1210 or 12.10%. 1.17 Illustrative Example 3: Find the area under the curve between z = -2.84 and z = 0.95. Step 1: Locate the area of the given z-value from the z-table. Standard Normal Probabletties Standard Normal Probabilities atle entry for : is the area under the standard normal curve tothe lot of t -30 6013 0013 0912 DOID .0019 0035 6034 0013 2 GSC-CID-LRMS-ESSLM, v.r. 03.00, Effective June 14, 2021 O2 GSC-CtD-LRMSESSLM, v.r. 02.00, Effective June 14, 2021 Step 2. Compute the area of the required region. 0.8266 P(2.B4 350) = P(Z > 2.25) = 1.0000 7 0.9373 = 0.0122 or 1.22% Step 3: Locate the area of the :- value from the z-table Step 4: Sketch the normal mirve and shade the desired area t 2 The probability that the volunteers will plant more than 850 trees per year is 1.22%. Step 5: State the conclustbn Illustrative Example 5: The sample average length of a sock that is produced at a factory is 34 cm and the sample standard deviation is 6. What is the probability that the length of a sock will be below 22 cm? 3 GSC-CtD-LRMSESSLM, v.r. 0300, Effective June 14, 2021 :.The area between 2 = 2.84 and z = 0.95. is 0.8266 or 82.66% . III. COMPUTING PROBABILITIES AND PERCENTILES USING THE STANDARD NORMAL TABLE Probabilities and percentiles can be solved using the standard normal table. The standard score is a measure of relative standing. It is a standard deviation value. It is the distance between the given value X and the mean. Any normal distribution can be transformed to a standard normal distribution and vice versa by the formula. Each of the values obtained using the given formula is called z-scores. Unknown Value Population Sample 1 F I i z-score z : z = a' 5 Score x=za+p xzs+i Mean p=xza i=xzs x _ _ - Standard Deviation a- = g F 5 = x x Z Illustrative Example 4: The number of mangrove trees planted by the volunteers is 679 per year with a standard deviation of T6. What is the probability that the volunteers will plant more than 850 trees per year? Step 1: Write the given and x = 850 I" : 6'79 unknown. 5 = 76 z =? Step2: Apply the formula. 2 = x _ 'u = m = 2.25 o' 76 PO: > 350) = 9(2 > 2.25) = 1.0000 0.9373 = 0.0122 or 1.22% Step 3: Locate the area of the 2- value from the z-table Step 4: Sketch the normal curve and shade the desired area D 13- I The probability that the volunteers will plant more than 850 trees per year is 1.22%. Step 5: State the conclusrbn Illustrative Example 5: The sample average length of a sock that is produced at a factory is 34 cm and the sample standard deviation is 6. What is the probability that the length of a sock will be below 22 cm? 3 GSC-CfD-LRMSESSLM, v.r. 0.100, Effectiveere 14, 2021 Step 1: Write the given and x = 22 I? = 34 unknown. 5 = 6 Z ='? Step2: Apply the formula. 2 = t3 = 22234 = th = 2 Step 3'. Locate the area of the 2- Hz