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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 OD 0,16​

Suppose A Normal Distribution Has A Mean Or 70 And A Standard Deviation Of 7 What Is Pxgt94 O A 0075 O 8 0025 C 084 OD 016 class=

Sagot :

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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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