Assignment 3-Q_ONLY (1).docx - BUSI 1013 \u2013 STATISTICS FOR BUSINESS ASSIGNMENT 3 STUDENT ID 2 UNIT NUMBER BUSI 1013 DATE OF EXAMINATION 0 2 1 8 0 8 1 1

# Assignment 3-Q_ONLY (1).docx - BUSI 1013 u2013 STATISTICS...

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BUSI 1013 – STATISTICS FOR BUSINESSASSIGNMENT # 3STUDENT ID2018081106UNIT NUMBER BUSI 1013UNIT CODEDATE OF EXAMINATION 271119Name (in Full) _____Gurleen kaur_____________________________________________________________________INSTRUCTIONS:Submission:There is no extension for submission or penalties for late submission. All assignments are to be summitted on the due date. No assignment should be submitted through email. In the event of an emergency preventing you from submitting on the deadline, special permission must be obtained from the Professor. Documentation substantiating emergency is required. In such a circumstance, if extension is granted, the Professor will open the submission function for you on an individual basis. DO NOT TEAR OUT ANY PART OF THIS SCRIPT BOOKQuestion #12345678910TotalExpected99912969101512Your marksTotal Marks: _______________/100
Answer all question and show steps in arriving at your answers. 1.Given that zis a StandardNormal random variable, find the following probabilities: (9 marks) 2.Given that z is a StandardNormal random variable, find z for each situation:(9 marks) For finding Z score in excel, Use the command = NORMSINV (0.68) = 0.47 (Rounded to the decimals) Z = 0.47
b.) We have given right side area = 0.68 We know that standard normal table always gives us left side area Left side area = 1 – 0.68 = 0.32 Z = NORMSINV (0.32) = -0.47c.)Total area to the left of -Z and right of Z is 0.32 Middle area -= 1 – 0.32 = 0.68 On each end we get = 0.32/2 = 0.16 Z = NORMSINV (0.16) = -0.99 We get -Z = -0.99 and Z = 0.99 3.Given that x is a Normal random variable with a mean of 10 and standard deviation of 4, findthe following probabilities:(9 marks) = P [Z < -0.825]= 1 – P [Z < 0.825] = 1 – 0.7967 P (x < 6.7) = 0.2033= P [Z > 0.625]= 1 – P [Z < 0.625]= 1 – 0.757P (x > 12.5) = 0.2643
= P [-0.3 < Z < 0.625]= P (ƶ< 0.625) – P (ƶ < -0.3) = 0.7357 – [1 – (P < 0.3)]= 0.7357 – [1 – 0.6179] = 0.7357 – [0.3821]P (8.8 < x < 12.5) = 0.35364.A simple random sample of 50 customers is selected from an account receivable portfolio and the sample mean account balance is \$1000. The population standard deviation σ is known to be \$200.(12 marks)a.Construct a 95% confidence interval for the mean account balance of the population.