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liubo4ka [24]
2 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]2 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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Asked on December 20, 2019 by

Ziya Vïvəķ

A cylindrical container with diameter of base 56cm contains sufficient water to submerge a rectangular solid of iron with dimensions (32×22cm×14cm). Find the rise in the level of water when the solid is completely submerged.

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ANSWER

Given,

Diameter of the base of the cylindrical container d=56 cm

Hence, radius of the base of the cylindrical container r=

2

56

=28 cm

Dimensions of rectangular solid is (32 cm×22 cm×14 cm)

We know that when a body is completely submerged in a container containing water,

the rise in the volume of water in the container is equal to the volume of the body.

Let, the water rises to a height h.

Hence,

Volume of water rises in cylindrical container =Volume of the rectangular solid

πr

2

h=32×22×14 cm

3

7

22

×(28)

2

×h=9856 cm

h=

22×784

9856×7

cm

h=4 cm

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