Respuesta :
Answer:
The probability that the sample mean is greater than 2.1 but less than 2.6 is 0.2236.
Step-by-step explanation:
The random variable X follows a discrete uniform distribution.
The probability mass function of X is:
[tex]f(x)=\left \{ {{\frac{1}{3}};\ x=1,2,3 \atop {0;\ \text{otherwise}}} \right.[/tex]
Then,
a = 1
b = 3
The mean and standard deviation of the random variable X are:
[tex]\mu=\frac{b+a}{2}=\frac{3+1}{2}=2\\\\\sigma=\sqrt{\frac{(b-a+1)^{2}-1}{12}}=\sqrt{\frac{(3-1+1)^{2}-1}{12}}=0.8165[/tex]
The sample size is, n = 39.
According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
Then, the mean of the sample means is given by,
[tex]\mu_{\bar x}=\mu[/tex]
And the standard deviation of the sample means is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]
As n = 39 > 30, the sampling distribution of sample mean of X will follow a Normal distribution approximately.
Compute the probability that the sample mean is greater than 2.1 but less than 2.6 as follows:
[tex]P(2.1<\bar X<2.6)=P(\frac{2.1-2.0}{0.8165/\sqrt{39}}<\frac{\bar X-\mu_{\bar x}}{\sigma/\sqrt{n}}<\frac{2.6-2.0}{0.8165/\sqrt{39}})[/tex]
[tex]=P(0.76<Z<4.59)\\\\=P(Z<4.59)-P(Z<0.76)\\\\=1-0.77637\\\\=0.22363\\\\\approx 0.2236[/tex]
Thus, the probability that the sample mean is greater than 2.1 but less than 2.6 is 0.2236.
