Question

The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent. If 500 births were observed rather than only 5, what is the approximate probability that at least 490 do not result in birth defects

Answer 1

Probability that at least 490 do not result in birth defects = 0.1076

Explanation:

Given - The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent.

To find - If 500 births were observed rather than only 5, what is the approximate probability that at least 490 do not result in birth defects

Proof -

Given that,

P(birth that result in a birth defect) = 1/33

P(birth that not result in a birth defect) = 1 - 1/33 = 32/33

Now,

Given that, n = 500

X = Number of birth that does not result in birth defects

Now,

P(X ≥ 490) =
`\sum\limits^(500)_(x=490) {^(500) C_(x) } ((32)/(33) )^(x) ((1)/(33) )^(500-x)`

=
`{^(500) C_(490) } ((32)/(33) )^(490) ((1)/(33) )^(500-490)` + .......+
`{^(500) C_(500) } ((32)/(33) )^(500) ((1)/(33) )^(500-500)`

= 0.04541 + ......+0.0000002079

= 0.1076

⇒Probability that at least 490 do not result in birth defects = 0.1076