Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 6.5, Problem 30E
To determine

To find:

The probability that out of 350 entering college freshmen, 100 will need to take developmental mathematics, if there is a 36% chance of a college freshman needing to take developmental mathematics.

Expert Solution & Answer
Check Mark

Answer to Problem 30E

Solution:

The probability that out of 350 entering college freshmen, 100 will need to take developmental mathematics is 0.0007.

Explanation of Solution

Given:

Total number of independent trials is 350.

Probability of an individual trial success is 36%.

Concept:

Continuity correction factor approximates discrete distributions to continuous ones by converting a random variable value of the former to a range of random variable values of the latter.

Binomial distribution is discrete and the probability of x successes out of n identical trials of the experiment is given by:

P(x)=Cnxpx(1p)nx,

With mean μ=np and standard deviation σ=np(1p),

Normal distribution is continuous and probability of obtaining the value z is expressed as:

f(z)=12πez22,

With mean 0 and standard deviation 1,

The continuity correction is done by changing the whole number x-value into an interval range:

[x0.5,x+0.5],

And then the probability P(xa) is approximated using the probability:

P(zza)=za12πez22dz,

Where za is the appropriate z-value.

In this question, discrete binomial distribution is approximated using the normal distribution, which is continuous.

Initially, the values of n and p are extracted after which the mean (μ) and the standard deviation (σ) are computed from binomial distribution.

Also, continuity correction is done, according to requirements of the question, that is by subtracting and adding 0.5 to the given x-value to make it interval.

Thereafter, the z-value is obtained using above x-value and the relation:

z=xμσ,

Following which, area under curve is computed by using a z-table.

Calculation:

Probability of a trial to be success:

p=36%=0.36

Probability of a trial to be failure:

q=1p=10.36=0.64

Then mean is:

μ=n×p=350×0.36=126

Also, standard deviation is:

σ=n×p×(1p)=μ×q=126×0.64=8.9799

Since here is exact “100”, continuity correction gives x-value as:

x1=99.5,x2=100.5.

This gives the z-value as:

z 1 = 99.5126 8.9799

    2.95

z 2 = 100.5126 8.9799

    2.84

Now, the required probability is:

=P(z22.84)P(z12.95)

=0.00230.0016

=0.0007

Conclusion:

The probability that out of 350 entering college freshmen, 100 will need to take developmental mathematics is 0.0007.

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Chapter 6 Solutions

Beginning Statistics, 2nd Edition

Ch. 6.1 - Prob. 11ECh. 6.1 - Prob. 12ECh. 6.1 - Prob. 13ECh. 6.1 - Prob. 14ECh. 6.1 - Prob. 15ECh. 6.1 - Prob. 16ECh. 6.1 - Prob. 17ECh. 6.1 - Prob. 18ECh. 6.1 - Prob. 19ECh. 6.1 - Prob. 20ECh. 6.1 - Prob. 21ECh. 6.1 - Prob. 22ECh. 6.1 - Prob. 23ECh. 6.1 - Prob. 24ECh. 6.1 - Prob. 25ECh. 6.2 - Prob. 1ECh. 6.2 - Prob. 2ECh. 6.2 - Prob. 3ECh. 6.2 - Prob. 4ECh. 6.2 - Prob. 5ECh. 6.2 - Prob. 6ECh. 6.2 - Prob. 7ECh. 6.2 - Prob. 8ECh. 6.2 - Prob. 9ECh. 6.2 - Prob. 10ECh. 6.2 - Prob. 11ECh. 6.2 - Prob. 12ECh. 6.2 - Prob. 13ECh. 6.2 - Prob. 14ECh. 6.2 - Prob. 15ECh. 6.2 - Prob. 16ECh. 6.2 - Prob. 17ECh. 6.2 - Prob. 18ECh. 6.2 - Prob. 19ECh. 6.2 - Prob. 20ECh. 6.2 - Prob. 21ECh. 6.2 - Prob. 22ECh. 6.2 - Prob. 23ECh. 6.2 - Prob. 24ECh. 6.2 - Prob. 25ECh. 6.2 - Prob. 26ECh. 6.2 - Prob. 27ECh. 6.2 - Prob. 28ECh. 6.2 - Prob. 29ECh. 6.2 - Prob. 30ECh. 6.2 - Prob. 31ECh. 6.2 - Prob. 32ECh. 6.2 - Prob. 33ECh. 6.2 - Prob. 34ECh. 6.2 - Prob. 35ECh. 6.2 - Prob. 36ECh. 6.2 - Prob. 37ECh. 6.2 - Prob. 38ECh. 6.2 - Prob. 39ECh. 6.2 - Prob. 40ECh. 6.2 - Prob. 41ECh. 6.2 - Prob. 42ECh. 6.2 - Prob. 43ECh. 6.2 - Prob. 44ECh. 6.2 - Prob. 45ECh. 6.2 - Prob. 46ECh. 6.2 - Prob. 47ECh. 6.2 - Prob. 48ECh. 6.2 - Prob. 49ECh. 6.3 - Prob. 1ECh. 6.3 - Prob. 2ECh. 6.3 - Prob. 3ECh. 6.3 - Prob. 4ECh. 6.3 - Prob. 5ECh. 6.3 - Prob. 6ECh. 6.3 - Prob. 7ECh. 6.3 - Prob. 8ECh. 6.3 - Prob. 9ECh. 6.3 - Prob. 10ECh. 6.3 - Prob. 11ECh. 6.4 - Prob. 1ECh. 6.4 - Prob. 2ECh. 6.4 - Prob. 3ECh. 6.4 - Prob. 4ECh. 6.4 - Prob. 5ECh. 6.4 - Prob. 6ECh. 6.4 - Prob. 7ECh. 6.4 - Prob. 8ECh. 6.4 - Prob. 9ECh. 6.4 - Prob. 10ECh. 6.4 - Prob. 11ECh. 6.4 - Prob. 12ECh. 6.4 - Prob. 13ECh. 6.4 - Prob. 14ECh. 6.4 - Prob. 15ECh. 6.4 - Prob. 16ECh. 6.4 - Prob. 17ECh. 6.4 - Prob. 18ECh. 6.4 - Prob. 19ECh. 6.4 - Prob. 20ECh. 6.4 - Prob. 21ECh. 6.4 - Prob. 22ECh. 6.4 - Prob. 23ECh. 6.4 - Prob. 24ECh. 6.4 - Prob. 25ECh. 6.4 - Prob. 26ECh. 6.4 - Prob. 27ECh. 6.4 - Prob. 28ECh. 6.4 - Prob. 29ECh. 6.5 - Prob. 1ECh. 6.5 - Prob. 2ECh. 6.5 - Prob. 3ECh. 6.5 - Prob. 4ECh. 6.5 - Prob. 5ECh. 6.5 - Prob. 6ECh. 6.5 - Prob. 7ECh. 6.5 - Prob. 8ECh. 6.5 - Prob. 9ECh. 6.5 - Prob. 10ECh. 6.5 - Prob. 11ECh. 6.5 - Prob. 12ECh. 6.5 - Prob. 13ECh. 6.5 - Prob. 14ECh. 6.5 - Prob. 15ECh. 6.5 - Prob. 16ECh. 6.5 - Prob. 17ECh. 6.5 - Prob. 18ECh. 6.5 - Prob. 19ECh. 6.5 - Prob. 20ECh. 6.5 - Prob. 21ECh. 6.5 - Prob. 22ECh. 6.5 - Prob. 23ECh. 6.5 - Prob. 24ECh. 6.5 - Prob. 25ECh. 6.5 - Prob. 26ECh. 6.5 - Prob. 27ECh. 6.5 - Prob. 28ECh. 6.5 - Prob. 29ECh. 6.5 - Prob. 30ECh. 6.CR - Prob. 1CRCh. 6.CR - Prob. 2CRCh. 6.CR - Prob. 3CRCh. 6.CR - Prob. 4CRCh. 6.CR - Prob. 5CRCh. 6.CR - Prob. 6CRCh. 6.CR - Prob. 7CRCh. 6.CR - Prob. 8CRCh. 6.CR - Prob. 9CRCh. 6.CR - Prob. 10CRCh. 6.CR - Prob. 11CRCh. 6.CR - Prob. 12CRCh. 6.CR - Prob. 13CRCh. 6.CR - Prob. 14CRCh. 6.CR - Prob. 15CRCh. 6.CR - Prob. 16CRCh. 6.CR - Prob. 17CRCh. 6.CR - Prob. 18CRCh. 6.CR - Prob. 19CRCh. 6.CR - Prob. 20CRCh. 6.CR - Prob. 21CRCh. 6.CR - Prob. 22CRCh. 6.P - Prob. 1PCh. 6.P - Prob. 2PCh. 6.P - Prob. 3PCh. 6.P - Prob. 4PCh. 6.P - Prob. 5PCh. 6.P - Prob. 6PCh. 6.P - Prob. 7PCh. 6.P - Prob. 8P
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