Suppose a normal distribution has a mean of 48 and a standard deviation of 2. What is the probability that a data value is betwe
en 43 and 46? Round your answer to the nearest tenth of a percent.
A. 17.2%
B. 16.3%
C. 14.2%
D. 15.2%
1 answer:
Answer:
Choice D. 15.2%
Step-by-step explanation:
We have a normal...
mean u = 48
standard deviation s = 2
We want P(43 < X < 46)
We standardize.
Consider P(43 < X) = P( (43 - 48)/2 < Z) = P(-2.5 < Z)
P( X < 46) = P( Z < (46 - 48)/2 ) = P(Z < -1)
We want P( -2.5 < Z < -1)
Look at Z-scores.
P( Z < -2.5) = 0.0062
P(Z < -1) = 0.1587
so P(-2.5 < Z < -1) = P(Z < -1) - P(Z < -2.5) = 0.1587 - 0.0062 = 0.1525 = 15.2%
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True. Math right to left would be absurd. Please Mark Brainliest!!!
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A=future amount
P=present amount
r=rate in decimal
t=time in years
to find the interest, we do A-P
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