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Sagot :
To find the Z-score for a giraffe that is 18.5 feet tall, we need to use the Z-score formula. The Z-score formula is a way of describing a value's relationship to the mean of a group of values. The formula is given by:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where:
- [tex]\( Z \)[/tex] is the Z-score,
- [tex]\( X \)[/tex] is the value we're evaluating,
- [tex]\( \mu \)[/tex] is the population mean,
- [tex]\( \sigma \)[/tex] is the population standard deviation.
Here’s the step-by-step process to find the Z-score:
1. Identify the given values:
- The height of the giraffe, [tex]\( X \)[/tex], is 18.5 feet.
- The population mean, [tex]\( \mu \)[/tex], is 16 feet.
- The population standard deviation, [tex]\( \sigma \)[/tex], is 1.8 feet.
2. Substitute the given values into the Z-score formula:
[tex]\[ Z = \frac{18.5 - 16}{1.8} \][/tex]
3. Perform the arithmetic operations:
- Calculate the numerator: [tex]\( 18.5 - 16 = 2.5 \)[/tex]
- Divide this result by the standard deviation: [tex]\( \frac{2.5}{1.8} \approx 1.3888888888888888 \)[/tex]
Hence, the Z-score for a giraffe that is 18.5 feet tall is approximately [tex]\( 1.3888888888888888 \)[/tex]. This means that the giraffe's height is about 1.39 standard deviations above the mean height of the population.
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where:
- [tex]\( Z \)[/tex] is the Z-score,
- [tex]\( X \)[/tex] is the value we're evaluating,
- [tex]\( \mu \)[/tex] is the population mean,
- [tex]\( \sigma \)[/tex] is the population standard deviation.
Here’s the step-by-step process to find the Z-score:
1. Identify the given values:
- The height of the giraffe, [tex]\( X \)[/tex], is 18.5 feet.
- The population mean, [tex]\( \mu \)[/tex], is 16 feet.
- The population standard deviation, [tex]\( \sigma \)[/tex], is 1.8 feet.
2. Substitute the given values into the Z-score formula:
[tex]\[ Z = \frac{18.5 - 16}{1.8} \][/tex]
3. Perform the arithmetic operations:
- Calculate the numerator: [tex]\( 18.5 - 16 = 2.5 \)[/tex]
- Divide this result by the standard deviation: [tex]\( \frac{2.5}{1.8} \approx 1.3888888888888888 \)[/tex]
Hence, the Z-score for a giraffe that is 18.5 feet tall is approximately [tex]\( 1.3888888888888888 \)[/tex]. This means that the giraffe's height is about 1.39 standard deviations above the mean height of the population.
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