Question

Recent research indicated that about 30​% of children in a certain country are deficient in vitamin D. A company that sells vitamin D supplements tests 330 elementary school children in one area of the country. Use a Normal approximation to find the probability that no more than 85 of them have vitamin D deficiency. The probability is approximately nothing. ​(Round to three decimal places as​ needed.)

Answer 1

Solution:

Given that,

P = 0.30

1 - P = 0.70

n = 330

Here, BIN ( n , P ) that is , BIN (330 , 0.30)

then,

n*p = 330*0.30 = 99 > 5

n(1- P) = 330*0.70 = 231 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 99

Standard deviation = =n*p*(1-p) = 330*0.30*0.70 = 69.3

We using countinuity correction factor

P( X a ) = P(X < a + 0.5)

P(x < 85.5) = P((x - ) / < (85.5 - 99) / 69.3)

= P(z < -1.62)

Probability = 0.053