Question

Let Y be a continuous random variable that is normally distributed with mean, µ = 30 and standard deviation, σ = 2.4, find P( 31.2 < y < 33.6) Answer should be accurate to 4 decimal places and entered in the format 0.XXXX

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To find P(31.2 < Y < 33.6) , we first standardize the values using the Z-score formula:

Z = \frac{X - \mu}{\sigma}

WhereX is the random variable, \mu is the mean, and \sigma is the standard deviation.

ForX = 31.2 :

Z_1 = \frac{31.2 - 30}{2.4} = \frac{1.2}{2.4} = 0.5

ForX = 33.6 :

Z_2 = \frac{33.6 - 30}{2.4} = \frac{3.6}{2.4} = 1.5

Now, we look up these Z-scores in the standard normal distribution table to find the corresponding probabilities:

P(0.5 < Z < 1.5)

Looking up in the standard normal distribution table:

P(0.5 < Z < 1.5) = P(Z < 1.5) - P(Z < 0.5)

P(Z < 1.5) = 0.9332

P(Z < 0.5) = 0.6915

P(0.5 < Z < 1.5) = 0.9332 - 0.6915 = 0.2417

Therefore,

P(31.2 < Y < 33.6) = 0.2417

\boxed{0.2417}

Where

For

For

Now, we look up these Z-scores in the standard normal distribution table to find the corresponding probabilities:

Looking up in the standard normal distribution table:

Therefore,

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