Question

Mariah Carl Pitman Sufolk Online Courses Library Content Collection Community Central Limit Theorem Practice: Module 7 The Central Limit Theorem e 7 The Central Limit Theorem Messages Redecm LatePass Due in 6 hours, 40 minutes, Due Sat 11/09/2019 11:59 pm 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 151000 dollars. Assume the standard deviation is 31000 dollars. Suppose you take a simple random sample of 80 graduates Find the probability that a single randomly selected policy has a mean value between 140948.9 and 156892 dollars P(140948.9 <X <156892) (Enter your answers as numbers accurate to 4 decimal places.) Find the probability that a random sample of size n 80 has a mean value between 140948.9 and 156892 dollars P(140948.9<M<156892) (Enter your answers as numbers accurate to 4 decimal places) Points possible: 1 Unlimited attempts Message instructor about this question Submit

Answer 1

Solution :

Given that ,

mean = $\mu$ = 151000

standard deviation = $\sigma$ = 31000

a.

P(140948.9 < x < 156892) = P[(140948.9 - 151000)/ 31000) < (x - $\mu$ ) /$\sigma$  < (156892 - 151000) / 31000) ]

= P(-0.32 < z < 0.19)

= P(z < 0.19) - P(z < -0.32)

= 0.5753 - 0.3745

= 0.2008

P(140948.9 < x < 156892) = 0.2008

b.

n = 80

$\mu$M = 151000

$\sigma$M = $\sigma$ / $\sqrt$ n = 31000 / $\sqrt$ 80 = 3465.9054

P(140948.9 < M < 156892) = P((140948.9 - 151000) / 3465.9054 < (M - $\mu$ M) / $\sigma$ M < (156892 - 151000) / 3465.9054)

= P(-2.90 < Z < 1.70)

= P(Z < 1.70) - P(Z < -2.90)

= 0.9554 - 0.0019

= 0.9535

P(140948.9 < M < 156892) = 0.9535