Tom scored better than 80.23% of the students who took the test and he will be admitted to this University.Īrea to the right (higher than) z = 1 is equal to 0.1586 = 15.87% scored more that 80.Īrea to the right of z = -1 is equal to 0.8413 = 84.13% should pass the test.Ĭ) 100% - 84.13% = 15.87% should fail the test.Īrea to the left (less than) of z = -0.5 is equal to 0.3085 = 30.85% earn less than $40,000.ī) For x = 45000, z = -0.25 and for x = 65000, z = 0.75Īrea between z = -0.25 and z = 0.75 is equal to 0.3720 = 37.20 earn between $45,000 and $65,000.Īrea to the right (higher) of z = 1 is equal to 0.1586 = 15.86% earn more than $70,000. A key tool for calculating proportions and percentages associated with normal distri- butions is the standard normal distribution. The proportion P of students who scored below 585 is given by If we multiply the values of the areas under the curve by 100, we obtain percentages.įor x = 585, z = (585 - 500) / 100 = 0.85 99.7 of the data points will fall within three standard deviations of the mean. 95 of the data points will fall within two standard deviations of the mean. Separate the lowest 40 from the rest of the distribution. The rule states that (approximately): - 68 of the data points will fall within one standard deviation of the mean. For a normal distribution, find the z-score that separates the distribution as follows: Separate the highest 30 from the rest of the distribution.52. The total area under the normal curve represents the total number of students who took the test. Find the range of values that defines the middle 80 of the distribution of SAT scores (372 and 628). x is normally ditsributed with a mean of 500 and a standard deviation of 100. 1.Introduction Normally distributed data sets are common in the real world. Thesenotes are self-contained and do not require you to have the textbook on hand. Let x be the random variable that represents the scores. Normal Distribution Worksheet This worksheet provides coverage ofnormally distributed data distributions. Use the 68-95-99.7 rule to find (a) the percentage of males with BMI less than 20.1. The probability that John's computer has a length of time between 50 and 70 hours is equal to 0.4082. The BMI for males age 20 to 74 is follows approximately a normal distribution with mean 27.9 and standard deviation 7.8. We have to find the probability that x is between 50 and 70 or P( 50< x < 70)įor x = 70, z = (70 - 50) / 15 = 1.33 (rounded to 2 decimal places) It has a mean of 50 and a standard deviation of 15. Let x be the random variable that represents the length of time. The probability that a car selected at a random has a speed greater than 100 km/hr is equal to 0.1587 X is a normally distributed variable with mean μ = 30 and standard deviation σ = 4. 3)Givenanormallydistributeddatasetwhosemeanis50andwhosestandarddeviationis10,what valueofwouldazscoreof2.5have 2.550 10 25. An online normal probability calculator and an inverse normal probability calculator may be useful to check your answers. The solutions to these problems are at the bottom of the page. Problems and applications on normal distributions are presented. Normal Distribution Problems with Solutions
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