Question

In the EAI sampling problem, the population mean is $51,900 and the population standard deviation is $4,000. When the sample size is n=20 there is a 0.4972 probability of obtaining a sample mean within ±$600 of the population mean. 

a. What is the probability that the sample mean is within $600 of the population mean if a sample of size 40 is used (to 4 decimals)?. 

b. What is the probability that the sample mean is within $600 of the population mean if a sample of size 80 is used (to 4 decimals)?

Answer 1

Solution :

Given that,

mean = $\mu$ = 51900

standard deviation = $\sigma$ = 4000

a)

n = 40

$\mu$_{$\bar x$} = $\mu$ = 51900

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 4000 / $\sqrt$ 40 = 632.4555

P( 51300 < $\bar x$ < 52500 ) = P((51300 - 51900) / 632.4555 <($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (52500-51900) / 632.4555))

= P(-0.95 < Z < 0.95)

= P(Z < 0.95) - P(Z < -0.95) Using z table,

= 0.8289- 0.1711

= 0.6578

b)

n = 80

$\mu$_{$\bar x$} = $\mu$ = 51900

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 4000 / $\sqrt$ 80 = 447.2136

P( 51300 < $\bar x$ < 52500 ) = P((51300 - 51900) / 447.2136<($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (52500-51900) / 447.2136))

= P(-1.34 < Z < 1.34)

= P(Z < 1.34) - P(Z < -1.34) Using z table,

= 0.9099 - 0.0901

= 0.8198