It is assumed that the test results for a class follow a normal distribution with a mean of 78 and a standard deviation of 36. If you know that a student's grade is greater than 72, what is the probability that it is greater than 84

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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

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