Answer to Question #118634 in Statistics and Probability for Michael

Question

Historically the results of a final exam given by a particular teacher have a mean of 65.3% with a standard deviation of 8.5

a. Calculate the probability of a student getting a mark between i. 75 and 85 ii. 79 and 81 iii. 79.9 and 80.1
b. Based on these results , a student says, “it’s almost impossible to get an 80 on the exam.” Comment on whether this is true

Answer 1

Historically the results of a final exam given by a particular teacher have a mean of 65.3% with a standard deviation of 8.5

a. Calculate the probability of a student getting a mark between i. 75 and 85 ii. 79 and 81 iii. 79.9 and 80.1

b. Based on these results , a student says, “it’s almost impossible to get an 80 on the exam.” Comment on whether this is true

Let X = random variable denoting the marks obtained by the student

When the distribution is not mentioned, it is conventionally assumed to be Normal.

Here the distribution of X has not been mentioned, so we assume

X ~ N($\mu$ = 65.3, $\sigma$² = 8.5²)

Then Z = $\frac{X-\mu}{\sigma}$ ~ N(0, 1), Z is called the standard normal variate

a.i. The probability of a student getting a mark between 75 and 85

= P(75 $\leq$ X $\leq$ 85)

= P($\frac{75-65.3}{8.5}\leq\frac{X-65.3}{8.5}\leq\frac{85-65.3}{8.5}$)

= P(1.14 $\leq$ Z $\leq$ 2.32)

= $\Phi$(2.32) - $\Phi$(1.14)

= 0.9898 - 0.8729 = 0.1169 [from standard normal distribution table]

ii. The probability of a student getting a mark between 79 and 81

= P(79 ≤ X ≤ 81)

= P($\frac{79-65.3}{8.5}\leq\frac{X-65.3}{8.5}\leq\frac{81-65.3}{8.5}$ ​)

= P(1.61 ≤ Z ≤ 1.85)

= $\Phi$(1.85) - $\Phi$(1.61)

= 0.9678 - 0.9463 = 0.0215 [from standard normal distribution table]

iii. The probability of a student getting a mark between 79.9 and 80.1

= P(79.9 ≤ X ≤ 80.1)

= P($\frac{79.9-65.3}{8.5}\leq\frac{X-65.3}{8.5}\leq\frac{80.1-65.3}{8.5}$)

= P(1.72 ≤ Z ≤ 1.74)

= $\Phi$(1.74) - $\Phi$(1.72)

= 0.9591 - 0.9573 = 0.0018 [from standard normal distribution table]

b. To test whether the claim of the student is true, we first find the probability of getting an 80 or more on the exam

= P(X $\geq$ 80)

= P($\frac{X-65.3}{8.5}\geq \frac{80-65.3}{8.5}$)

= P(Z $\geq$ 1.73)

= 1 - P(Z < 1.73)

= 1 - $\Phi$(1.73)

= 1 - 0.9582 = 0.0418 = 4.2% (approx)

Answer: a. The probability of the student getting a mark between i. 75 and 85 is 0.1169 ii. 79 and 81 is 0.0215 iii. 79.9 and 80.1 is 0.0018.

b. As the probability of getting an 80 or more is 4.2%, we can say that the claim by the student is not entirely true although the chances are low.