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Wittaler [7]
3 years ago
13

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 )​
Mathematics
1 answer:
Gelneren [198K]3 years ago
5 0

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.

<em><u>Let </u></em>\bar X<em><u> = sample mean</u></em>

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 = <u>0.0287</u>

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.

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