ETF5900 Business Statistics Tutorial 5 The normal distribution Part A (Complete BEFORE tutorial and submit online. Use Excel where necessary and write answers on question sheet) A5.1. Instructions for the use of the normal distribution functions NORM.S.DIST, NORM.S.INV, NORM.DIST, and NORM.INV are provided in section E4.4 of the Excel notes in the Software Information Folder under Week 0 on Moodle. (a) Use the Excel function NORM.S.DIST to calculate the following probabilities to four decimal places, where the random variable Z follows a standard normal distribution. Write your answers to four decimal places on this sheet and remember to draw the curves. For every second question, please use your tables to find the probability as well as Excel. (i) P( Z < 0.443) = (ii) P (Z <1.522) = (iii) P( Z > 1.944) = (iv) P(0 < Z < 1.282) = (v) P (1.522 < Z <1.958) = (vi) P(-1 < Z < 1.282) = 1 ETF5900 Business Statistics Tutorial 5 Now we begin with a given probability and have to work backwards to find the corresponding z value. (b) Use the Excel function NORM.S.INV, and draw curves, to find the value z* for which (i) (ii) P( Z < z* ) = 0.895 (iii) P( Z < z* ) = 0.054 (iv) P( Z > z* ) = 0.025 (v) P(0 < Z < z* ) = 0.400 (vi) (c) P( Z < z* ) = 0.5 P( z* < Z < z* ) = 0.882 Let X be a normally distributed random variable with mean 10 and standard deviation 4.5. (i) Use the Excel function NORM.DIST to find P( X < 0) . 2 ETF5900 Business Statistics (ii) Tutorial 5 Use the Excel function NORM.INV to find the value of x* such that P( X < x* ) = . 0.10 A5.2 Suppose Z has a normal distribution with mean 0 and standard deviation 1. Use a diagram in working out the following. (a) Use NORM.S.DIST to find the area of the upper tail cut off by 2.327 in the standard normal distribution. (b) Use NORM.S.INV to find the value of z0.002 , i.e. the z value that cuts off an upper tail of area 0.002. (c) Use NORM.S.INV to find the two values of z that cut off two tails of total area 0.002. 3 ETF5900 Business Statistics Tutorial 5 Part B: To be completed in class B5.3. (a) When using tables to obtain standard normal probabilities, values of Z can only be specified to two decimal places. Use Table 1a or 1b in the statistical tables provided on the website to obtain the probabilities you already calculated in A5.1. You will need to round the z-values to two decimal places in order to use the tables, and you will therefore observe some differences from the answers obtained using Excel. (i) P( Z < 0.443) (ii) P (Z <1.522) (iii) P( Z > 1.944) (iv) P (0 < Z <1.282) (v) P(1.522 < Z <1.958) (vi) P(-1 z* ) = 0.025 (v) P(0 < Z < z* ) = 0.400 (vi) B5.4. P( z* < Z < z* ) = 0.882 The lifetimes of the heating element in a Heatfast electric oven are normally distributed, with a mean of 7.8 years and a standard deviation of 2.0 years. (a) (i) If the element is guaranteed for 2 years, what percentage of the ovens sold will need replacement in the guarantee period because of element failure? Include a statement describing your answer. 5 ETF5900 Business Statistics (ii) Tutorial 5 In a year in which 10,000 ovens are sold, how many ovens would you expect to have to replace in the guarantee period because of element failure? (b) What proportion of elements are expected to last for between 2 and 10 years? Write a statement explaining your answer. (c) Heatfast is reconsidering the length of the guarantee period on heating elements. Calculate the length of guarantee period such that Heatfast would expect to replace a maximum of 1% of ovens due to element failure. Include a statement describing your answer. (d) Find the length of time such that it includes 95% of all ovens. Include a statement describing your answer. 6 ETF5900 Business Statistics Tutorial 5 B5.5. Based on laboratory testing, the lifetime of a Tyrannosaurus brand tyre is taken to be Normally distributed, with mean 65,107 kilometres (km) and standard deviation of 2,582 km. The tyres carry a customer warranty for 60,000 km. (a) What proportion of the tyres is expected to fail before the warranty expires? (b) Obtain, the 50th percentile of tyre life. Explain how you obtained this answer. (c) Explain, to a non-statistician, what this value means. (d) Probability that a tyre would last 145,000 km or more. 7