A set of data has a mean of 56.1. The data follows a normal distribution curve and has a standard deviation of 8.2. Find the pro bability that a randomly selected value is greater than 67.5 A.
0.0823
B.
1.39
C.
0.9177
D.
–1.08
2 answers:
Given that mean=56.1 and standard deviation=8.2, P(x>67.5) will be found as follows: The z-score is given by: z=(x-μ)/σ thus the z-score will be given by: z=(67.5-56.1)/8.2 z=11.4/8.2 z=1.39 thus P(z=1.39)=0.9177 thus: P(x>67.5)=1-P(z>0.9177) =1-0.9177 =0.0823 Answer: A. 0.0823
Answer: A. 0.0823
Step-by-step explanation:
Given : Mean =56.1
Standard deviation =8.2
P(x>67.5) will be found as follows:
The z-score is given by:
Substitute the values of means ans standard deviation in it, we get
As
P(z<1.39)=0.9177
thus:
P(X>67.5)=1-P(z<1.39)
⇒P(X>67.5)=1-0.9177
∴P(X>67.5)=0.0823
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