The United States Department of Agriculture (USDA) found tha

stop2dance3l

stop2dance3l

Answered question

2021-12-12

The United States Department of Agriculture (USDA) found that the proportion of young adults ages 20-39 who regularly skip eating breakfast is 0.2380.238. Suppose that Lance, a nutritionist, surveys the dietary habits of a random sample of size ?=500n=500 of young adults ages 20-39 in the United States.
Apply the central limit theorem to find the probability that the number of individuals, ?,X, in Lances

Answer & Explanation

Heather Fulton

Heather Fulton

Beginner2021-12-13Added 31 answers

Step 1
Population proportion: P=0.238
Sample size =n=500
a) To find probability that number of individuals =X>126
By Using Central Limit theorem,
XN(mean=nP=5000.238=119,SD=5000.238(10.238)=9.5225
P(X>126)=P(X1199.5225>1261199.5225)=0.2311
Step 2
b) To find probability that number of individuals =X<98
By Using Central Limit theorem,
XN(mean=nP=5000.238=119,SD=5000.238(10.238)=9.5225
P(X>98)=P(X1199.5225>981199.5225)=0.01371
Using Excel,
Mean=nP=119variance=90.678=5000.238(10.238)SD=9.5225P(X>126)=0.231139=1NORMDIST(126,B1,B3,1)P(X<98)=0.013716=NORMDIST(98,B1,B3,1)
porschomcl

porschomcl

Beginner2021-12-14Added 28 answers

Step 1
This is a case where we will use Normal Approximation to Binomial Distribution.
Let X be random variable. So we know:
XB(n, p)
Then, Normal Approximation would be:
XNormal Approx(np, npq)
Given in the problem, sample n=500 and probability of skipping, p=0.238, so:
X(500, 0.238)
Note: q=1p=10.238=0.762
Thus,
XNormal Approx(119, 90.678)
Now,
We need P(X>122).
We convert to Z by using formula:
z=xμσ
Where μ is the mean, or np and σ is the standard deviation, which is npq
Thus, we have:
P(X>122)
=P(Xμσ>12211990.678)
=P(Z>0.37)
Which can be said as:
1P(Z<0.37)
Using Z-Table (normal table), we have:
10.6443=0.3557
This is our answer.

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