Question

Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean

= 28.3 kilograms and standard deviation
= 4.0 kilograms. Let x be the weight of a fawn in kilograms. Convert the following x intervals to z intervals. (Round your answers to two decimal places.)

(a) x < 30

(b) 19 < x

(c) 32 < x < 3

Answer 1

To convert weights to z-scores for a normal distribution, the formula z = (x - μ) / σ is used. For example, for a fawn weighing less than 30 kilograms, the z-score is 0.42. However, there appears to be a typo in the question for the interval 32 < x < 3.

Step-by-step explanation:

The question is about converting the weight of a fawn from a normal distribution to z-scores. A z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. To convert an x value to a z-score, we use the formula z = (x - μ) / σ, where μ is the mean, σ is the standard deviation, and x is the weight of the fawn.

1. For x < 30, the z-score would be z = (30 - 28.3) / 4.0 = 0.42.

2. For 19 < x, converting 19 to a z-score yields z = (19 - 28.3) / 4.0 = -2.33.

3. Finally, for 32 < x < 3, this appears to be a typo, as it does not make sense that x would be less than 3 and greater than 32 simultaneously. Assuming it meant 32 < x, the z-score for x = 32 would be z = (32 - 28.3) / 4.0 = 0.93.