Calculate the Probability of a Normally Distributed Random Sample

Question

Please i would like to understand these problems about probability distributions, I can't find a right solution for this problem.

I have a variable X which is the level of glucose in blood and is normally distributed, I also have the mean μ = 200 mg/dl and the standard deviation σ = 20mG/dl.

The treaky question is: Find the probability that from 100 random samples, at least 10 will have at least 178 level of glucose in blood?

What I tried to do was:

1. P(X<=178)
2. P(X<=178) = P(X−μ<=178−200 )=P((X−μ)/σ<=(178−200)/20)
3. P(X<=178) = P(Z<=−1.1) = 0.1357 (for 1 person)
4. 0.1357 * 10 = 1.357 for 10 persones

I think this is logical But it's incorrect i don't know what am i doing wrong.

note: tha data is not real it's just an example

Answer 1

let $A_i$ indicate that sample $i$ had at least $178$ level of glucose. The probability $\mathbb{P}\{A_i\}$ can be calculated. Denote it by $p$.

We now need to find the probability that out of $A_i,i=1...100$ independent variables, $10$ will be true. So we have $100$ Bernoulli variables with success probability $p$. This is the Binomial distribution with $N=100,P=p$.

We want the probability that $10/100$ are successful. Denote $A$ as this Binomial variable. We want the probability: $P(A>=10)=1-P(A<10)$, and this can be calculated as a function of the Binomial cdf or by summing the pdf: $$ 1-\sum_{k=1}^9 {N \choose k} p^k(1-p)^{N-k}$$.