Respuesta :
Answer:
The probability is 0.789
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. 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 pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\sigma = 16, n = 400, s = \frac{16}{\sqrt{400}} = 0.8[/tex]
Find the probability that the mean age of the individuals sampled is within one year of the mean age for all individuals in the city.
Using [tex]\mu = 30[/tex], this is the pvalue of Z when X = 31 subtracted by the pvalue of Z when X = 29. So
X = 31
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{31 - 30}{0.8}[/tex]
[tex]Z = 1.25[/tex]
[tex]Z = 1.25[/tex] has a pvalue of 0.8944
X = 29
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{29 - 30}{0.8}[/tex]
[tex]Z = -1.25[/tex]
[tex]Z = -1.25[/tex] has a pvalue of 0.1056
0.8944 - 0.1056 = 0.7888
Rounded to three decimal places, 0.789
The probability is 0.789
Answer:
Step-by-step explanation:
Hello!
To investigate the relationship between age and annual medical expenses a random sample of 400 individuals was taken.
The sample is expected to be representative of the entire population.
a.
The variable of interest is X: the age of an individual in the city.
The standard deviation is known to be δ= 16
Applying the Central Limit Theorem we can approximate the distribution of the sampling distribution to normal.
Using this approximation you can use the standard normal distribution [tex]Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }[/tex]≈N(0;1)
P(μ-1≤X[bar]≤μ+1)= P(X[bar]≤μ+1)-P(X[bar]≤μ-1)
Now using the standard normal distribution you have to standardize both terms:
P(X[bar]≤μ+1)= P(Z≤[(μ+1)-μ]/(σ/√n))= P(Z≤(μ-μ+1)/(15/√400))= P(Z≤1/(15/√400))= P(Z≤1.33)
P(X[bar]≤μ-1)= P(Z≤[(μ-1)-μ]/(σ/√n))= P(Z≤(μ-μ-1)/(15/√400))= P(Z≤-1/(15/√400))= P(Z≤-1.33)
As you see there is no need to know the actual value of the population mean since you have to subtract it when calculating the Z values.
P(Z≤1.33) - P(Z≤-1.33)= 0.90824 - 0.09176= 0.81648
I hope it helps!
