Answer: 0.4337
Step-by-step explanation:
Let X represents the test results for a class that follow a normal distribution .
Given: Mean [tex]\mu=78[/tex], Standard deviation [tex]\sigma=36[/tex]
Then, the probability that it is greater than 84 will be
[tex]P(X>84)=P(\dfrac{X-\mu}{\sigma}>\dfrac{84-78}{36})\\\\=P(Z>0.167)\ \ \ [Z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Z<0.167)\\\\=1-0.5663=0.4337\ [\text{By p-value table}][/tex]
Hence, the required probability = 0.4337