Answer:
1: 18.75
Step-by-step explanation:
We multiply both sides by 5.
Answer:
1899
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 3234
Standard deviation = 871
Percentage of newborns who weighed between 1492 grams and 4976 grams:
1492 = 3234 - 2*871
So 1492 is two standard deviations below the mean.
4976 = 3234 + 2*871
So 4976 is two standard deviations above the mean.
By the Empirical Rule, 95% of newborns weighed between 1492 grams and 4976 grams.
Out of 1999:
0.95*1999 = 1899
So the answer is 1899
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:
1 2/3 = 5/3
6 3/4 = 27/4
-9 7/3 = -34/3
4 2/7 = 30/7
Step-by-step explanation:
Answer:
15%
Step-by-step explanation:
34.50-30=4.50
10% of 30= 3
5%of 30 = 1.50
3+1.50=4.50
so 4.50 = 15%