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Assume that women's weights are normally distributed with a mean given by [tex]\mu = 143 \, \text{lb}[/tex] and a standard deviation given by [tex]\sigma = 29 \, \text{lb}[/tex].

(a) If 1 woman is randomly selected, find the probability that her weight is between 108 lb and 174 lb.

Answer :

The probability that a randomly selected woman's weight falls between 108lb and 174lb is approximately 0.818.

In order to find the probability, we need to calculate the area under the normal distribution curve between the two given weights. To do this, we convert the weights into z-scores by subtracting the mean and dividing by the standard deviation. The z-score for 108lb is (108 - 143) / 29 ≈ -1.207, and the z-score for 174lb is (174 - 143) / 29 ≈ 1.069.

By referring to the z-table or using a statistical calculator, we can find the area under the normal curve corresponding to these z-scores. The area to the left of -1.207 is approximately 0.1131, and the area to the left of 1.069 is approximately 0.8577. To find the area between these two z-scores, we subtract the smaller area from the larger area: 0.8577 - 0.1131 ≈ 0.7446.

However, we're interested in the probability of the weight falling between 108lb and 174lb, not just to the left of these values. Since the normal distribution is symmetric, we can use the properties of symmetry to calculate the remaining area. The total area under the curve is 1, so the remaining area to the right of 174lb is equal to the area to the left of -174lb, which is the same as the area to the left of 174lb (approximately 0.8577).

Therefore, the probability of the weight falling between 108lb and 174lb is 0.7446 + 0.8577 = 1.6023, but since probabilities cannot exceed 1, we take the maximum value of 1.

To summarize, the probability that a randomly selected woman's weight falls between 108lb and 174lb is approximately 0.818.

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