Question

Birthweights at a local hospital have a normal distribution with a mean of 110.3 oz. and a variance of 82.1 ounces squared. Calculate the proportion of infants with birthweights either below 106 oz. or above 120 oz.

Answer 1

Solution:

Given, the Normal distribution with,

Mean   = 110.3

Variance  ² = 82.1

So ,

Standard deviation = 82.1 = 9.06090503206

P(either below 106 oz. or above 120 oz.)

= 1 - { Between 106 and 120}

= 1 - { P(106 < X < 120) }

= 1 - { P(X < 120) - P(X < 106) }

= 1 - { P[(X - )/ <  (120 - 110.3)/9.06090503206  - P[(X - )/ <  (106 - 110.3)/9.06090503206 }

= 1 - { P(Z < 1.07) - P(Z < -0.47) }

= 1 - {0.8577 - 0.3192}

= 0.4615