Answer:
C
Step-by-step explanation:
Cuz I said so
Answer:
1/4 hopefully this helps you with work
Answer:
0.7486 = 74.86% observations would be less than 5.79
Step-by-step explanation:
I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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.
The general format of the normal distribution is:
N(mean, standard deviation)
Which means that:
![\mu = 5.43, \sigma = 0.54](https://tex.z-dn.net/?f=%5Cmu%20%3D%205.43%2C%20%5Csigma%20%3D%200.54)
What proportion of observations would be less than 5.79?
This is the pvalue of Z when X = 5.79. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{5.79 - 5.43}{0.54}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B5.79%20-%205.43%7D%7B0.54%7D)
![Z = 0.67](https://tex.z-dn.net/?f=Z%20%3D%200.67)
has a pvalue of 0.7486
0.7486 = 74.86% observations would be less than 5.79
Answer:
It is 2 hours, 22 minutes, 48 seconds.
The 2 is obviously 2 hours. For the minutes, .38 * 60 = 22.8, so the minutes is 22. 0.8 * 60 = 48. So 2 hours, 22 minutes and 48 seconds.
Answer:
x = 32
Step-by-step explanation: