Suppose a normal distribution has a mean of 79 and a standard deviation of7. What is P(x86)?A. 0.84B. 0.16C. 0.975O D. 0.025

SOLUTION
Step 1 :
In this question, we are meant to find the probability of the normal distribution, given that
the mean = 79 , standard deviation = 7 .
Step 2 :
Using the Normal Distribution, we have that :
[tex]\begin{gathered} Z\text{ = }\frac{X\text{ - }\mu\text{ }}{\sigma} \\ \text{where X = 86, }\mu\text{ = 79, }\sigma\text{ = 7} \end{gathered}[/tex][tex]\begin{gathered} P(X\ge86) \\ \\ Z\text{ =}\frac{\text{ 86 - 79}}{7} \\ \text{ Z =}\frac{7}{7} \\ Z\text{ =1} \end{gathered}[/tex]Step 3 :
We need to find the area under the normal distribution curve.
From the Normal Distribution Curve, Z = 1 lies in the 84. 1 %
But we need to find Probability ( greater than 1 ) = 100 % - 84. 1 % = 15. 9 %
Step 4 :
We need to simplify 15.9 % to decimal =
[tex]\frac{15.\text{ 9}}{100}\text{ = 0.159 = 0.16 ( 2 decimal places )}[/tex]CONCLUSION :
The final answer is 0.16 - OPTION B