Question

2. (14 pts) A population of 1,000 students spends an average of $10.50 a day on dinner. The standard deviation of the expenditure is $3. A simple random sample of 64 students is taken. a. What are the expected value, standard deviation, and shape of the sampling distribution of the sample mean? Explain the reason. Which theorem ensures the shape of the sampling distribution? b. What is the probability that these 64 students will spend a combined total of more than $715.21? c. What is the probability that these 64 students will spend a combined total between $703.59 and $728.45?

Answer 1

(1) Eppeded Value z -= 10.50 standard Dawation of Sampleoso como 0.375 Sampling distribution of sample mean Normally distilbud - DS 0-5 3164 b) combind spend = 15.2 Individual spende #15:21 2 11.175156 : * = 11,175156 Now, p (>11.175155) > p (Try this . PETS2 - 10-5) 2p(2719) e 1-P (261.8) e 1-0-9641 ::P 17711-1772) = 0.6359 c) combined spend a 103.45 a score = 10.9936-100 Insivited spend a go-aa26 combined spend 2 728.45 2 1.31628 Insivited Spendz 11:3820 z score = 11:3820-10. Now Pl10.9936CX 511: 3829 = P(1.3162562423521) 7169 101 31564 . 22.3521 P (242.3521)–p (261.51623 obtaind from std normal table 20.990666-0.905955 +10.8470) |