Question

Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 170000 dollars. Assume the standard deviation is 33000 dollars. Suppose you take a simple random sample of 97 graduates.

Find the probability that a single randomly selected policy has a mean value between 168994.8 and 178041.5 dollars.

P(168994.8 < X < 178041.5) = ____________ (Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a random sample of size n = 97 has a mean value between 168994.8 and 178041.5 dollars.

P(168994.8 < M < 178041.5) = ___________ (Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

Solution! u=170000 6- 33000 n= af Mi= 170000 GX-6 = 33000 = 3350.6423 o Jaz PC168994.8 că 5 178041.5) =P 1689948-170000 Lines5178041.5-170000 L3350.6423 6x 3350.6923 =P -1,005 2 3350.6423 <z s 804105 3250.6423 =P (-0.30 < 25 2640] =P [ 2<2.40] -p [z(-0.30] = 0.9918 – 0.382) - 0.6097 P(168994.8<x< 1780415)= 0.6097 probability – 0-6097

um=170000 om = G = 33000 = 33506423 97 PC 168994.8 cm < (78041.5) P 168494.8-170000 SM-UM<17804105-170000 I 3350.6423 6M 3350.6423 =P -1.oos.2 3350.6423 a z c 8041.5 33 50:6423 -PT-0.30 azc2.907 =P [222.40J-o [Z<-0.30] = 0.9918 -0. 3821 =0.6097 P(168494.8 CM < 1780415) = 0.6097 probability - 0.6097