Respuesta :
Answer:
a) Approximately [tex]-1.05[/tex].
b) Approximately [tex]2.49[/tex].
c) Approximately [tex]0.13[/tex].
d) Approximately [tex](25.6, \, 51.0)[/tex].
e) Approximately [tex](12.9,\, 63.7)[/tex].
f) Approximately [tex](21.8,\, 54.8)[/tex].
Step-by-step explanation:
The [tex]z[/tex]-score of a measurement [tex]x[/tex] gives the number of standard deviations [tex]\sigma[/tex] between that measurement and the mean [tex]\mu[/tex] of the dataset.
For example, for this dataset, it is given that [tex]\mu = 38.3[/tex] whereas [tex]\sigma = 12.71[/tex]. For a [tex]25[/tex]-year-old viewer, [tex]x = 25[/tex].
How far away is this observation from the mean of this population? [tex](x - \mu) = 25 - 38.3 = -13.3[/tex].
The [tex]z[/tex]-score of this observation would be the approximate number of standard deviations that corresponds to this difference:
[tex]\displaystyle z = \frac{x - \mu}{\sigma} = \frac{25 - 38.3}{12.71} \approx -1.05[/tex].
Similarly:
For [tex]x = 70[/tex], [tex]\displaystyle z = \frac{x - \mu}{\sigma} = \frac{70 - 38.3}{12.71} \approx 2.49[/tex].
For [tex]x = 40[/tex], [tex]\displaystyle z = \frac{x - \mu}{\sigma} = \frac{40 - 38.3}{12.71} \approx 0.13[/tex].
Separately adding and subtracting [tex]\sigma[/tex] from [tex]\mu[/tex] would give the upper bound and the lower bound of the age range within [tex]1[/tex] standard deviation from the mean:
- Upper bound: [tex]\mu + \sigma = 38.3 + 12.71 \approx 51.0[/tex].
- Lower bound: [tex]\mu - \sigma = 38.3 - 12.71 \approx 25.6[/tex].
Similarly, for the interval within [tex]2[/tex] standard deviations from the mean:
- Upper bound: [tex]\mu + 2\, \sigma = 38.3 + 2 \times 12.71 \approx 63.7[/tex].
- Lower bound: [tex]\mu - 2\, \sigma = 38.3 - 2\times 12.71 \approx 12.9[/tex].
For the interval within [tex]1.3[/tex] standard deviations from the mean:
- Upper bound: [tex]\mu + 1.3\, \sigma = 38.3 + 1.3 \times 12.71 \approx 54.8[/tex].
- Lower bound: [tex]\mu - 2\, \sigma = 38.3 - 2\times 12.71 \approx 21.8[/tex].