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Chapter 6: Problem 33

Let \(x\) be a continuous random variable that is normally distributed with a mean of 25 and a standard deviation of 6 . Find the probability that \(x\) assumes a value a. between 29 and 36 b. between 22 and 35

Short Answer

Expert verified
The probability for part (a) is 0.2178 while for part (b) it is 0.6440.

Step by step solution

01

Convert to standard normal distribution (Z-Score)

Transform the given ranges into standard normal using the Z-score formula. For part a. \(Z = \frac{(29 - 25)}{6}\), and \(Z1 = \frac{(36-25)}{6}\)For part b. \(Z2 = \frac{(22-25)}{6}\), and \(Z3 = \frac{(35-25)}{6}\)
02

Use the z-table

Refer to the z-table (also called standard normal table) to find the probabilities corresponding to z-scores calculated in step 1. Additionally, the z-table provides the probability that a random variable stemming from a standard normal distribution is lesser than z.For the intervals given in part (a),The probability for \(Z = 0.67\) (approximated) and \(Z1 = 1.83\) (approximated) are about 0.7486 and 0.9664 respectively.For the intervals given in part (b),The probability for \(Z2 = -0.5\) (approximated) and \(Z3 = 1.67\) (approximated) are about 0.3085 and 0.9525 respectively.
03

Compute the probability in the given range

To find the probability that the variable lies within the said range, subtract the smaller probability from the larger one.For part a, the probability \(P(29 < x < 36)\) is \(0.9664 - 0.7486 = 0.2178\).For part b, the probability \(P(22 < x < 35)\) is \(0.9525 - 0.3085 = 0.6440\).

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