Question

A population has a mean of 200 and a standard deviation of 70. Suppose a sample of size 100 is selected and is used to estimate . Use z-tableWhat is the probability that the sample mean will be within +/- 6 of the population mean (to 4 decimals)What is the probability that the sample mean will be within +/- 19 of the population mean

Answer 1

Solution :

Given that,

mean = = 200

standard deviation = = 70

n = 100

= = 200

= / n = 70 / 100 = 7

a) P(194 < < 206)  

= P[(194 - 200) / 7 < ( - ) / < (206 - 200) / 7)]

= P( -0.86 < Z < 0.86)

= P(Z < 0.86) - P(Z < -0.86)

Using z table,  

= 0.8051 - 0.1949   

= 0.6102

b) P(181 < < 219)  

= P[(181 - 200) / 7 < ( - ) / < (219 - 200) / 7)]

= P( -2.71 < Z < 2.71)

= P(Z < 2.71) - P(Z < -2.71)

Using z table,  

= 0.9966 - 0.0034  

= 0.9932