3.  Suppose the random variable X𝑋 is normally distributed with μ=50𝜇=50 and σ2=36𝜎2=36 . What is the value of b𝑏 such that P(X≥b)=0.3264P(𝑋≥𝑏)=0.3264 ?Multiple choice 1 Question 3  45.8  59.1  52.7  47.3

To solve this problem, we need to use the properties of the standard normal distribution.

1. First, we need to find the z-score that corresponds to the given probability. The z-score is a measure of how many standard deviations an element is from the mean. In this case, we are given P(X≥b)=0.3264. This is the same as saying P(Z≥z)=0.3264, where Z is a standard normal random variable.

2. We can look up this probability in a standard normal distribution table, or use a calculator or software that can do this. The z-score that corresponds to a right-tail probability of 0.3264 is approximately -0.45. This is because the total area under the curve of a standard normal distribution is 1, so the area to the left of the z-score is 1 - 0.3264 = 0.6736. The z-score that corresponds to this left-tail probability is -0.45.

3. Now that we have the z-score, we can find the corresponding value of b. The formula to convert a z-score to an original score (b) is b = μ + zσ, where μ is the mean and σ is the standard deviation. In this case, μ = 50 and σ = sqrt(36) = 6.

4. Substituting these values into the formula gives b = 50 + (-0.45)*6 = 47.3.

So, the value of b such that P(X≥b)=0.3264 is 47.3. Therefore, the correct answer is 47.3.

To solve this problem, we need to use the properties of the standard normal distribution.

1. First, we need to find the z-score that corresponds to the given probability. The z-score is a measure of how many standard deviations an element is from the mean. In this case, we are given P(X≥b)=0.3264. This is the same as saying P(Z≥z)=0.3264, where Z is a standard normal random variable.

2. We can look up this probability in a standard normal distribution table, or use a calculator or software that can do this. The z-score that corresponds to a right-tail probability of 0.3264 is approximately -0.45.

3. Now we can use the formula for transforming a z-score in a standard normal distribution to a value in our specific normal distribution. The formula is X = μ + zσ. Plugging in the given values, we get b = 50 + (-0.45)sqrt(36) = 50 - 0.456 = 47.3.

So, the value of b such that P(X≥b)=0.3264 is 47.3.