Answer:
a) 0.0721
Step-by-step explanation:
Let the random variable X be the score on the ACT exam.
Given:
- mean (μ) = 21
- standard deviation (σ) = 5
Therefore, the random variable X is normally distributed:
[tex]X \sim\textsf{N}(21,5^2)[/tex]
Z-score for a sample mean
[tex]\textsf{If }\: X \sim\textsf{N}(\mu,\sigma^2)\:\textsf{ then }\: \overline{X} \sim \left(\mu,\dfrac{\sigma^2}{n}\right) \implies Z=\dfrac{\overline{X}-\mu}{\sigma / \sqrt{n}} \sim \textsf{N}(0,1)[/tex]
[tex]\textsf{then the test statistic is}: \quad z=\dfrac{\overline{x}-\mu}{\sigma / \sqrt{n}}[/tex]
Given:
- [tex]\overline{x}[/tex] = 19.8
- μ = 21
- σ = 5
- n = 37
Therefore, the test statistic is:
[tex]\implies z=\dfrac{19.8-21}{5/ \sqrt{37}}=-\dfrac{6\sqrt{37}}{25}=-1.459863007=-1.46\:\: \sf (3\:s.f.)[/tex]
Using the Z-tables to find the probability:
[tex]\implies \text{P}(Z \leq -1.46)=0.0721[/tex]
