Question

A population has a mean of 300 and a standard deviation of 90. Suppose a sample of size 100 is selected and is used to estimate . Use z-table. What is the probability that the sample mean will be within +/- 3 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.) What is the probability that the sample mean will be within +/- 19 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

Answer 1

Solution :

Given that,

mean = = 300

standard deviation = = 90

n = 100

= = 300

= / n = 90 / 100 = 9

P(297 < < 303) = P((297 - 300) / 9<( - ) / < (303 - 300) / 9))

= P( -0.33< Z < 0.33)

= P(Z < 0.33) - P(Z < -0.33) Using standard normal table,  

= 0.6293 - 0.3707

Probability = 0.2586

b)

  

= P( -2.11< Z < 2.11)

= P(Z < 2.11) - P(Z < -2.11) Using standard normal table,  

= 0.9826 - 0.0174

Probability = 0.9652