Answer :
We begin by noting that the final exam scores are normally distributed with a mean of [tex]$\mu = 70$[/tex] and a standard deviation of [tex]$\sigma = 4$[/tex]. To determine Jonah's standing in percentiles, we first calculate the [tex]$z$[/tex]-score corresponding to his score of 78. The [tex]$z$[/tex]-score is given by
[tex]$$
z = \frac{x - \mu}{\sigma},
$$[/tex]
where [tex]$x$[/tex] represents Jonah's score. Substituting the given values, we have
[tex]$$
z = \frac{78 - 70}{4} = \frac{8}{4} = 2.
$$[/tex]
Next, we use the standard normal distribution to find the cumulative probability corresponding to [tex]$z = 2$[/tex]. This cumulative probability tells us the fraction of students who scored below Jonah's score. For [tex]$z = 2$[/tex], the cumulative probability is approximately 0.97725. Converting this into a percentage, we get
[tex]$$
0.97725 \times 100 \approx 97.7\%.
$$[/tex]
Thus, Jonah's score of 78 places him in the 97.7th percentile.
The correct answer is hence [tex]$\boxed{97.7\%}$[/tex].
[tex]$$
z = \frac{x - \mu}{\sigma},
$$[/tex]
where [tex]$x$[/tex] represents Jonah's score. Substituting the given values, we have
[tex]$$
z = \frac{78 - 70}{4} = \frac{8}{4} = 2.
$$[/tex]
Next, we use the standard normal distribution to find the cumulative probability corresponding to [tex]$z = 2$[/tex]. This cumulative probability tells us the fraction of students who scored below Jonah's score. For [tex]$z = 2$[/tex], the cumulative probability is approximately 0.97725. Converting this into a percentage, we get
[tex]$$
0.97725 \times 100 \approx 97.7\%.
$$[/tex]
Thus, Jonah's score of 78 places him in the 97.7th percentile.
The correct answer is hence [tex]$\boxed{97.7\%}$[/tex].