In: Statistics and Probability
Recent research indicated that about 20 % of children in a certain country are deficient in vitamin D. A company that sells vitamin D supplements tests 370 elementary school children in one area of the country. Use a Normal approximation to find the probability that no more than 59 of them have vitamin D deficiency.
Solution:
Given that,
P = 0.20
1 - P = 0.80
n = 370
Here, BIN ( n , P ) that is , BIN (370 , 0.20)
then,
n*p = 370*0.20 = 74 > 5
n(1- P) = 370*0.80 = 296 > 5
According to normal approximation binomial,
X Normal
Mean = = n*P = 74
Standard deviation = =n*p*(1-p) = 370*0.20*0.80= 59.2
We using countinuity correction factor
P( X a ) = P(X < a + 0.5)
P(x < 59.5) = P((x - ) / < (59.5 - 74) / 59.2)
= P(z < -1.88)
= 0.0301
The probability that no more than 59 of them have vitamin D deficiency is 0.0301