Anyone plz help me ty
2 answers:
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Answer:
A customer who sends 78 messages per day would be at 99.38th percentile.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
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 p-value, we get the probability that the value of the measure is greater than X.
Average of 48 texts per day with a standard deviation of 12.
This means that
a. A customer who sends 78 messages per day would correspond to what percentile?
The percentile is the p-value of Z when X = 78. So
has a p-value of 0.9938.
0.9938*100% = 99.38%.
A customer who sends 78 messages per day would be at 99.38th percentile.
Answer:
<em>Terry had £780 initially and Faye had £1300</em>
Step-by-step explanation:
<u>System of Equations</u>
We'll call the following variables:
x = Terry's balance in his bank account
y = Faye's balance in her bank account
The initial relation between them is:
Cross-multiplying:
5x = 3y [1]
If Terry put £220 in his account he had x+220 and if Faye withdrew £300 from her account, she had y-300. Both quantities are equal, thus
x + 220 = y - 300
Subtracting 220:
x = y - 520 [2]
Substituting in [1]
5(y - 520) = 3y
Multiplying:
5y - 2600 = 3y
Adding 2600 and subtracting 3y:
2y = 2600
Dividing by 2:
y = 1300
From [2]:
x = 1300 - 520
x = 780
Terry had £780 initially and Faye had £1300