Question

compute the z - score corresponding to each of the values of x below.
a. ( x = 28 ), ( s = 4 ), ( overline{x}=20 ) b. ( x = 88 ), ( mu = 87 ), ( sigma = 2 ) c. ( mu = 45 ), ( sigma = 5 ), ( x = 45 )
d. ( s = 6 ), ( x = 35 ), ( overline{x}=50 ) e. in parts a - d, state whether the z - score locates x within a sample or a population. f. in parts a - d, state whether each value of x lies above or below the mean and by how many standard deviations.
e. for part a, detern
the z - score locates
for part b, determi
the z - score locates
for part c, determir
the z - score locates
for part d, determi
the z - score locates
f. for part a, state t
the value of x lies

Explanation:

Step1: Recall the z - score formula

For a sample: \(z=\frac{x - \bar{x}}{s}\), for a population: \(z=\frac{x-\mu}{\sigma}\)

Step2: Calculate z - scores for each part

Part a

Given \(x = 28\), \(s=4\), \(\bar{x}=20\) (sample). Using \(z=\frac{x - \bar{x}}{s}\), we have \(z=\frac{28 - 20}{4}=\frac{8}{4}=2\)

Part b

Given \(x = 88\), \(\mu = 87\), \(\sigma=2\) (population). Using \(z=\frac{x-\mu}{\sigma}\), we have \(z=\frac{88 - 87}{2}=\frac{1}{2}=0.5\)

Part c

Given \(x = 45\), \(\mu = 45\), \(\sigma=5\) (population). Using \(z=\frac{x-\mu}{\sigma}\), we have \(z=\frac{45 - 45}{5}=0\)

Part d

Given \(x = 35\), \(s = 6\), \(\bar{x}=50\) (sample). Using \(z=\frac{x - \bar{x}}{s}\), we have \(z=\frac{35 - 50}{6}=\frac{- 15}{6}=-2.5\) (assuming there was a typo in the problem statement as per the formula application)

Step3: Determine sample/population for each part

• Part a: Since we use \(\bar{x}\) (sample mean) and \(s\) (sample standard deviation), \(x\) is in a sample.
• Part b: Since we use \(\mu\) (population mean) and \(\sigma\) (population standard deviation), \(x\) is in a population.
• Part c: Since we use \(\mu\) (population mean) and \(\sigma\) (population standard deviation), \(x\) is in a population.
• Part d: Since we use \(\bar{x}\) (sample mean) and \(s\) (sample standard deviation), \(x\) is in a sample.

Step4: Determine position relative to the mean

• Part a: \(z = 2>0\), so \(x\) is \(2\) standard deviations above the mean.
• Part b: \(z=0.5>0\), so \(x\) is \(0.5\) standard deviations above the mean.
• Part c: \(z = 0\), so \(x\) is at the mean.
• Part d: \(z=-2.5<0\), so \(x\) is \(2.5\) standard deviations below the mean.

Answer:

• a: \(z = 2\), sample, \(2\) standard deviations above the mean.
• b: \(z = 0.5\), population, \(0.5\) standard deviations above the mean.
• c: \(z = 0\), population, at the mean.
• d: \(z=-2.5\), sample, \(2.5\) standard deviations below the mean.