Question

Suppose babies born in a large hospital have a mean weight of 3366 grams, and a variance of 244,036. If 118 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by less than 45 grams? Round your answer to four decimal places.

Answer 1

Solution :

Given that ,

mean = = 3366

variance = ² = 244,036

standard deviation = = ² = 244,036 = 494

n = 118

=   = 3366

= / n = 494 / 118 = 45.48

3366  ± 45 = 3321, 3411

P(3321 < < 3411)  

= P[(3321 - 3366) / 45.48 < ( - ) / < (3411 - 3366) / 45.48)]

= P( -0.99 < Z < 0.99)

= P(Z < 0.99) - P(Z < -0.99)

Using z table,  

= 0.8389 - 0.1611

= 0.6778