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liubo4ka [24]
3 years ago
9

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?
Mathematics
1 answer:
zaharov [31]3 years ago
8 0
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.

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madreJ [45]

Answer:

The total length of rebar used is  21.66 meters.

Step-by-step explanation:

Given:

Ivan had cut a reinforcing bar in 19 pieces and length of each bar is 1.14 meters.

Number of pieces = 19

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We need to find the total length of reinforcing bar.

To calculate the total length we will multiply number of pieces with length of each piece.

Hence,

Total length of rebar = Number of pieces × Length of each piece = 19\times1.14 = 21..66 m

Rounding to nearest hundred = 21.66 m

Hence the total length of reinforcing bar is 21.66 meters.

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