Suppose a normal distribution has a mean or 70 and a standard deviation of 7. What is P(x>=94)? O A. 0.075 O 8. 0.025 c. 0.84
1 answer:
Answer:
0.025
Step-by-step explanation:
Mean = 79 ; Standard deviation = 7
P(x ≥ 93)
We obtain the ZSCORE :
Zscore = (x - mean) / standard deviation
x = 93
Zscore = (93 - 79) / 7 = 2
P(Z ≥ 2) = 0.02275
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