Respuesta :
Given Information:
Mean percent of childhood asthma = μ = 2.23%
Sample size = n = 31
Standard deviation = σ = 1.24%
Required Information:
P(X > 2.6) = ?
Answer:
P(X > 2.6) = 0.0475 = 4.75%
Step-by-step explanation:
Let X is the random variable that represents percent of childhood asthma prevalence.
We want to find out the probability that the mean childhood asthma prevalence for the sample is greater than 2.6%
[tex]P(X > 2.6) = 1 - P(X < 2.6)\\\\P(X > 2.6) = 1 - P(Z < \frac{x - \mu}{\frac{\sigma}{\sqrt{n} } } )\\\\P(X > 2.6) = 1 - P(Z < \frac{2.6 - 2.23}{\frac{1.24}{\sqrt{31} } } )\\\\P(X > 2.6) = 1 - P(Z < \frac{2.6 - 2.23}{0.222} )\\\\P(X > 2.6) = 1 - P(Z < \frac{0.37}{0.222} )\\\\P(X > 2.6) = 1 - P(Z < 1.67)\\[/tex]
The z-score corresponding to 1.67 is 0.9525
P(X > 2.6) = 1 - 0.9525
P(X > 2.6) = 0.0475
P(X > 2.6) = 4.75%
Therefore, there is 4.75% probability that the mean childhood asthma prevalence for the sample is greater than 2.6%
How to use z-table?
Step 1:
In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.6)
Step 2:
Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 1.67 then go for 0.07 column)
Step 3:
Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.
Using the normal distribution and the central limit theorem, it is found that that there is a:
0.0485 = 4.85% probability that the mean childhood asthma prevalence for the sample is greater than 2.6%. The interpretation is that when many samples of 31 are chosen, around 4.85% of them will have an asthma prevalence greater than 2.6%.
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When the distribution is normal, the z-score formula is used, with [tex]\mu[/tex] as the mean and [tex]\sigma[/tex] as the standard deviation.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The Z-score measures how many standard deviations the measure is from the mean.
- Each z-score has a p-value associated with it.
- This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
- Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
- For a sample of n observations, by the Central Limit Theorem, the standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
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- Mean of 2.23% means that [tex]\mu = 2.23[/tex]
- Standard deviation of 1.24% means that [tex]\sigma = 1.24[/tex]
- Sample of 31 means that [tex]n = 31, s = \frac{1.24}{\sqrt{31}} = 0.2227[/tex]
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The probability of a sample mean above 2.6% is 1 subtracted by the p-value of Z when X = 2.6. Thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2.6 - 2.23}{0.2227}[/tex]
[tex]Z = 1.66[/tex]
[tex]Z = 1.66[/tex] has a p-value of 0.9515.
1 - 0.9515 = 0.0485.
0.0485 = 4.85% probability that the mean childhood asthma prevalence for the sample is greater than 2.6%. The interpretation is that when many samples of 31 are chosen, around 4.85% of them will have an asthma prevalence greater than 2.6%.
A similar problem is given at https://brainly.com/question/22934264
