Question

A simple random sample of size equals 49 is obtained from a population with mu equals 84 and sigma equals 14. ​(a) Describe the sampling distribution of x overbar. ​(b) What is Upper P (x overbar greater than 87.8 )​

Answer 1

Answer:

(a) The sampling distribution of $\bar X$ is given by;

             $\bar X$ ~ Normal ($\mu =84, s = \frac{\sigma}{\sqrt{n} } =\frac{14}{\sqrt{49} }$ )

(b) P($\bar X$ > 87.8) = 0.0287

Step-by-step explanation:

We are given that a simple random sample of size equals 49 is obtained from a population with mu equals 84 and sigma equals 14.

Let $\bar X$ = sample mean

The z-score probability distribution for sample mean is given by;

                             Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 84

            $\sigma$ = standard deviation = 14

            n = sample size = 49

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

(a) The sampling distribution of $\bar X$ is given by;

             $\bar X$ ~ Normal ($\mu =84, s = \frac{\sigma}{\sqrt{n} } =\frac{14}{\sqrt{49} }$ )

(b) Probability of $\bar X$ greater than 87.8 is given by = P($\bar X$ > 87.8)

     

      P($\bar X$ > 87.8) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{87.8-84}{\frac{14}{\sqrt{49} } }$ ) = P(Z > 1.90) = 1 - P(Z $\leq$ 1.90)

                                                        = 1 - 0.9713 = 0.0287

The above probability is calculated by looking at the value of x = 1.90 in the z table which has an area of 0.9713.