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Answer:
The z score for your flight's taxi time is of 2.14.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
An airline claims that its average taxi time is 15 minutes, and the standard deviation is 1.4 minutes.
This means that [tex]\mu = 15, \sigma = 1.4[/tex]
On a flight with this airline, you observe that the taxi time is 18 minutes. Calculate the z score for your flight's taxi time.
This is Z when [tex]X = 18[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{18 - 15}{1.4}[/tex]
[tex]Z = 2.14[/tex]
The z score for your flight's taxi time is of 2.14.
You can convert the variate which tracks heights of trees to standard normal variate. That value of standard normal variate is called the z score and shows the area of the considered area in the plot of the standard normal distribution.
The z score for our flight's taxi time is approx 2.14286
How to convert a normal distribution variate to standard normal distribution variate?
Suppose you have got a normal distribution variate with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] (let the variate is denoted by X), then we have
[tex]X \sim N(\mu, \sigma)[/tex]
To convert it to standard normal distribution with mean 0 and the standard deviation as 1 ( [tex]Z \sim N(0,1)[/tex] ), we have:
[tex]Z = \dfrac{X - \mu}{\sigma}[/tex]
For the given case we have:
Let the average taxi time be tracked by random variable X, then, as the mean time is 15 minutes, and the standard deviation 1.4 minutes , thus,
[tex]X \sim N(15, 1.4)[/tex]
Converting it to standard variate:
[tex]Z = \dfrac{X - 15}{1.4}[/tex]
Since the observed taxi time was 18 minutes, thus, we have X = 18,
Getting the z score:
[tex]Z = \dfrac{18 - 15}{1.4} = \dfrac{3}{1.4} \approx 2.14286[/tex]
Thus,
The z score for our flight's taxi time is approx 2.14286
Learn more here about normal distribution here:
brainly.com/question/14989264
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