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The confidence interval is from 9.81 to 10.19.
We first find the mean of the data:
<span>(9.8+10.2+10.4+9.8+10.0+10.2+9.6)/7 = 10
Next we find the standard deviation:
</span>σ=√([<span>(9.8-10)^2+(10.2-10)^2+(10.4-10)^2+(9.8-10)^2+(10-10)^2+(10.2-10)^2+(9.6-10)^2]/7) = 0.262
The z-score for 95% confidence is found by
1-0.95 = 0.05; 0.05/2 = 0.025; from the z-table, it is 1.96.
The confidence interval is calculated using
</span>
<span>
</span>
We first find the mean of the data:
<span>(9.8+10.2+10.4+9.8+10.0+10.2+9.6)/7 = 10
Next we find the standard deviation:
</span>σ=√([<span>(9.8-10)^2+(10.2-10)^2+(10.4-10)^2+(9.8-10)^2+(10-10)^2+(10.2-10)^2+(9.6-10)^2]/7) = 0.262
The z-score for 95% confidence is found by
1-0.95 = 0.05; 0.05/2 = 0.025; from the z-table, it is 1.96.
The confidence interval is calculated using
</span>
</span>
Answer:
nasa pic yung answer
Step-by-step explanation:
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Answer:
The probability the event that none of them are obese is 0.0352.
Step-by-step explanation:
<h3>Binomial distribution:</h3>A discrete random variable X having to the set {0,1,2,3....,n} as the spectrum, is said to have binomial distribution with parameters n= the number of trial, and p= probability of successes on an individual trial , if the p.m.f of X is given by,
for x=0,1,2,...,n
=0 elsewhere.
where 0<p<1 and n an positive integer,
Given that,
The probability of the event that a child living in an urban area in the united state is obese is 20%.
n=Number of children = 15, p= 20%= 0.20.
The probability the event that none of them are obese is
=P(X=0)
=0.0352