Pediatric asthma survey, n = 50. suppose that asthma affects 1 in 20 children in a population. you take an srs of 50 children fr
om this population. can the normal approximation to the binomial be applied under these conditions? if not, what probability model can be used to describe the sampling variability of the number of asthmatics?
In this situation, n=50, p=1/20, q=(1-p)=19/20, and npq=19/8=2.4
We would like np and npq to be a large number, at least greater than 10. The normal approximation can always be applied, but the result will be very approximate, depending on the values of np and npq.
Situations are favourable for the normal approximation when p is around 0.5, say between 0.3 and 0.7, and n>30.
"Normal approximation" is using normal probability distribution to approximate the binomial distribution, when n is large (greater than 70) or exceeds the capacity of most hand-held calculators. The binomial distribution can be used if the following conditions are met: 1. Bernoulli trials, i.e. exactly two possible outcomes.2. Number of trials is known before and constant throughout the experiment, i.e. independent of outcomes.3. All trials are independent of each other.4. Probability of success is known, and remain constant throughout trials. If all criteria are satisfied, we can model with binomial distribution, where the probability of x successes out of N trials each with probability of success p is given byP(x)=C(N,x)(p^x)(1-p)^(N-x)and,C(N,x) is number of combinations of selecting x objects out of N. The mean is np, and variance is npq.
For the given situation, np=2.5, npq=2.375, so standard deviation=sqrt(2.375)=1.54.
331.3% Convert 53/16 to simplest form, 3 5/16, convert 5/16 into numbers —> 0.3125 Take the whole number and add it into the ones place: 3.3125 Move the decimal into two places then round to the nearest tenth: 331.3
remember you can do anything to an equaiton as long as you do it to boh sides ad try to get unknown to one side 9n-10=4n+10 minus 4n both sides 9n-4n-10=4n-4n+10 5n-10=0+10 add 10 5n+10-10=0+10+10 5n+0=20 5n=20 divide both sies by 5 5n/5=20/5 5/5n=4 1n=4 n=4