For problems 1-7, use the following problem description. The true average weight
ID: 3065362 • Letter: F
Question
For problems 1-7, use the following problem description. The true average weight of a 20-yr-old MTSU male student will be assumed to be 170 pounds with a standard deviation of 24 pounds 1. If we take a random sample of 36 20-yr-old MTSU males and calculate their average weight, what is the probability that the sample mean is between 162 pounds and 178 pounds? 2. Between what two values (in pounds) will 80% of the sample means of size 36 lie? 3. Between what two values (in pounds) will 90% of the sample means of size 36 lie? For Z- 1.645 we obtain (170-(1.645)(24/6), 170 +(1.645)(24/6) (163.42, 176.58) 4. Between what two values (in pounds) will 95% of the sample means of size 36 lie? For Z-1.96 we obtain (170 (1.96)X24/6),170 (1.96) (24/6)) (162.16, 177.84) 5. Between what two values (in pounds) will 98% of the sample means of size 36 lie? For Z-2.33 we obtain (170-(2.33)(24/6), 170 (2.33) 24/6)) (160.68, 179.32) 6. Between what two values (in pounds) will 99% of the sample means of size 36 lie? For Z-2.575 we obtain (170 (2.575)(24/6), 170 +(2.575)(24/6) (159.7,180.3) 7. Therefore, 99% of the sample means of size 36 lie within what "distance in pounds" of the population mean of 170 pounds? Using Z-2.575 we obtain 2.575(24/6) 10.3 pounds. Hence, within a "distance' of 10.3 pounds on either side of the population mean pounds on either side of the population mean we will find 99% ofall sample means of size 36. Page 1 of 2Explanation / Answer
mean = 170
std. dev. = 24
#1.
P(162 < X < 178)
= P((162 - 170)/(24/sqrt(36) < z < (178 - 170)/(24/sqrt(36))
= P(-2 < z < 2)
= P(z < 2) - P(z < -2)
= 0.9772 - 0.0228 .. (using standard z table)
= 0.9545
#2.
SE = sd / sqrt(n) = 24/sqrt(36) = 4
z -value = +/- 1.28
xbar = 170 - 1.28*4 = 164.88
xbar = 170 + 1.28*4 = 175.12
Hence values are 164.88, 175.12
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