A RANDom variable X is normally distributed with mean 20 and standard deviation 4. find: a) P(X>25) B)P(X<25) C)The value of d such that P(20<X<D)=0.4641 D)P(15 "LESS THAN OR EQUAL TO" X "LESS THAN OR EQUAL TO 20"
Question
A RANDom variable X is normally distributed with mean 20 and standard deviation 4. find: a) P(X>25) B)P(X<25) C)The value of d such that P(20<X<D)=0.4641 D)P(15 "LESS THAN OR EQUAL TO" X "LESS THAN OR EQUAL TO 20"
Solution
a) To find P(X>25), we first need to convert the X value to a z-score using the formula z = (X - μ) / σ, where μ is the mean and σ is the standard deviation. So, z = (25 - 20) / 4 = 1.25. Looking up this z-score in the standard normal distribution table, we find that the area to the left of this z-score is 0.8944. Since we want the probability that X is greater than 25, we need to find the area to the right of the z-score, which is 1 - 0.8944 = 0.1056.
b) To find P(X<25), we use the same z-score calculated in part a. The probability that X is less than 25 is the same as the area to the left of the z-score, which is 0.8944.
c) To find the value of d such that P(20<X<D)=0.4641, we first find the z-score that corresponds to the given probability. Looking up 0.4641 in the standard normal distribution table, we find a z-score of approximately 0.05. We then convert this z-score back to an X value using the formula X = zσ + μ, so d = 0.05*4 + 20 = 20.2.
d) To find P(15 ≤ X ≤ 20), we first convert the X values to z-scores. For X = 15, z = (15 - 20) / 4 = -1.25. For X = 20, z = (20 - 20) / 4 = 0. Looking up these z-scores in the standard normal distribution table, we find that the area to the left of z = -1.25 is 0.1056 and the area to the left of z = 0 is 0.5. The probability that X is between 15 and 20 is the difference between these two areas, which is 0.5 - 0.1056 = 0.3944.
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