The lengths of a particular snake are approximately normally distributed with a given mean mc025-1.jpg = 15 in. and standard dev
iation mc025-2.jpg = 0.8 in. What percentage of the snakes are longer than 16.6 in.?
0.3%
2.5%
3.5%
5%
0.3%
2.5%
3.5%
5%
2 answers:
Answer:
2.5% of the snakes exceed the length of 16.6 inches
Detailed explanation:
The Z score is calculated as follows:
Wherein:
X = raw score = 16.6
μ = mean = 15
σ = standard deviation = 0.8
<pconsequently>The Z-score signifies how far the data diverges from the mean in terms of standard deviations. Hence, the value 16.6 lies 2 standard deviations above the mean.
About 95% of all data falls within two standard deviations from the mean. (spanning from mean-2 times the standard deviation to mean+2 times the standard deviation)
Thus, 5% of the total data is outside this range. To determine how many snakes exceed 16.6 inches, we must identify the area to the right of this length.
This remaining 5% is evenly split into two parts.
Consequently, of the snakes surpass 16.6 inches
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To determine if there is evidence suggesting a change in average height, we can conduct a right-tailed test and formulate both null and alternative hypotheses.
H₀ (null hypothesis): μ = 162.5
H₁ (alternative hypothesis): μ > 162.5
With two samples to analyze, we can calculate the z-score using the formula provided below.
In this formula, Z symbolizes the z-score, Χ denotes the new sample mean, μ indicates the theoretical average, δ represents the standard deviation, and n signifies the sample size. Based on the gathered values,
Assuming a significance level of α = 0.05. With a z-score of 2.77, we can reference the z-table to ascertain the p-value. This yields P(Z > 2.77) =.0028. Since our p-value is below α, we reject the null hypothesis, indicating that the average height of female freshman students has indeed shifted.
H₀ (null hypothesis): μ = 162.5
H₁ (alternative hypothesis): μ > 162.5
With two samples to analyze, we can calculate the z-score using the formula provided below.
In this formula, Z symbolizes the z-score, Χ denotes the new sample mean, μ indicates the theoretical average, δ represents the standard deviation, and n signifies the sample size. Based on the gathered values,
Assuming a significance level of α = 0.05. With a z-score of 2.77, we can reference the z-table to ascertain the p-value. This yields P(Z > 2.77) =.0028. Since our p-value is below α, we reject the null hypothesis, indicating that the average height of female freshman students has indeed shifted.