Suppose babies born in a large hospital have a mean weight of 3831 grams, and a variance of 108,900. If 65 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

the probability that the mean weight of the sample babies would differ from the population mean by less than 45 grams is 0.7284

Step-by-step explanation:

Given that

mean weight = 3831 grams,

variance = 108,900

sample size n = 65

the probability that the mean weight of the sample babies would differ from the population mean by less than 45 grams = ?

Now standard error of mean = √( 108900 / 65 ) = 40.9314

z = +/- (45/40.9314)

z = +/-1.0994

so

P( -1.0994 <z< 1.0994 ) = P( z<1.0994 ) - (p<-1.0994 )

= 0.8642 - 0.1358

= 0.7284

Therefore the probability that the mean weight of the sample babies would differ from the population mean by less than 45 grams is 0.7284

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the 14th row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100