The lengths of infants at birth in a certain hospital are normally distributed with a mean of 18 inches and a standard deviation of 2.2 inches. find the probability that an infant selected at random from this hospital measures between 16 and 21 inches.

Mean, m = 18 in
Standard deviation, SD = 2.2 in
Range: 16 ≤ X ≤ 21 in

Calculating Z value,
Z = (X-m)/SD

Then,
Z1 = (16-18)/2.2 ≈ -0.91
Z2 = (21-18)/2.2 ≈ 1.36
From Z table, and at Z1 = -0.91, and Z2 = 1.36;
P(16) = 0.1814
P(21) = 0.9131

Therefore,
P(16≤X≤21) = 0.9131 - 0.1814 = 0.7317
The probability that a child selected randomly measures between 16 and 21 in is 0.7317.

Value is 6 i think k