There are 35 boys in sixth grade. The number of girls in the sixth grade is 42. Lonnie says that means the ratio of number of bo
2 answers:
Answer:
No.
Step-by-step explanation:
Ratio of any objects can be calculated by dividing the two objects with the common number.
Boys in the sixth grade = 35.
Girls in the sixth grade = 42.
Ratio of boys to girl =
Ratio =
Ratio = 5:6. (the numbers are divided by 7).
Thus, the ratio is 5:6 and not the 5:7.
Thus, Lonnie is not correct.
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Answer:
0.227 = 22.7% probability that the mean printing speed of the sample is greater than 18.12 ppm.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean of 17.42 ppm and a standard deviation of 3.25 ppm.
This means that
Sample of 12:
This means that
Find the probability that the mean printing speed of the sample is greater than 18.12 ppm.
This is 1 subtracted by the p-value of Z when X = 18.12.
By the Central Limit Theorem
has a pvalue of 0.773.
1 - 0.773 = 0.227
0.227 = 22.7% probability that the mean printing speed of the sample is greater than 18.12 ppm.