Question

Suppose x has a distribution with a mean of 90 and a standard deviation of 21. Random samples of size n = 36 are drawn. (a) Describe the x distribution and compute the mean and standard deviation of the distribution has ---Select- distribution with meanz - and standard deviation o, - (b) Find the z value corresponding to x = 83. ZE (c) Find P(x < 83), (Round your answer to four decimal places.) P(x < 83) = (d) Would it be unusual for a random sample of size 36 from the x distribution to have a sample mean less than 83? Explain. No, it would not be unusual because less than 5% of all such samples have means less than 83. Yes, it would be unusual because less than 5% of all such samples have means less than 83. Yes, it would be unusual because more than 5% of all such samples have means less than 83. No, it would not be unusual because more than 5% of all such samples have means less than 83.

Answer 1

solution Coiven_thein formation Meanu = 90.1 Standaod deviation=6=2 0 - 36 . Here follows the NOGOL distoibution, because according f_ceniaal lift theorem tom X follolas. Moomal distoibution with mean-UkQ standaod deviation -_67= Mean=42 = 90, 67= 26 - 3-5_ V36 :Ă has nogmal distoibution with mean 4ă901 | and standard deviatios &_=3.5] 10 Z-value is zeg 4 OR Z-44 Z83-90 . 3-5 D IE-2 ): PCX 283) = P(XEM < 8 ) Hallording to Central Jint thatod theodem - =P2483.99) 2010 ZX-4 = P(Z (-2) ---P67283) = 0.02287- (From z-table) .. pa <83) = 0·0228 OR 2.28 % @ONO. it wowd not be unsual because 1933 than 5% ofart such somptes Tove Dean 123 THT) 89. Reeuse we get poo bability that Aysamplo novcan test than 83 is 2.28% which is 1955 than 5%. Pron@7 correef