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wlad13 [49]
3 years ago
5

Suppose that Upper X has a discrete uniform distribution f left-parenthesis x right-parenthesis equals StartLayout left-brace1st

Row 1st Column 1 divided by 3, 2nd Column x equals 1,2,3 2nd Row 1st Column 0, 2nd Column otherwise EndLayout A random sample of n equals 39 is selected from this population. Find the probability that the sample mean is greater than 2.1 but less than 2.6. Express the final answer to four decimal places (e.g. 0.9876). The probability is
Mathematics
1 answer:
DiKsa [7]3 years ago
6 0

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 <em>X</em> follows a discrete uniform distribution.

The probability mass function of <em>X</em> is:

f(x)=\left \{ {{\frac{1}{3}};\ x=1,2,3 \atop {0;\ \text{otherwise}}} \right.

Then,

a = 1

b = 3

The mean and standard deviation of the random variable <em>X</em> are:

\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

The sample size is, <em>n</em> = 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,

\mu_{\bar x}=\mu

And the standard deviation of the sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

As <em>n</em> = 39 > 30, the sampling distribution of sample mean of <em>X</em> 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:

P(2.1

                          =P(0.76

Thus, the probability that the sample mean is greater than 2.1 but less than 2.6 is 0.2236.

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