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Answer:
Yes, he earned a Varsity letter because he played in 61% of the games.
Step-by-step explanation:
To know the percent of games he played out of the total, we can do it by dividing the games he played by the total of games and multiply this by 100% to get the percent of games he played:
We solve:
0.611111*100% = 61.11%
So we know that he earned a letter because he played in 61.11% of the games, more than he needed.
Answer:
a) A sample of 256 was used in this survey.
b) 45.14% probability that the point estimate was within ±15 of the population mean
Step-by-step explanation:
This question is solved using the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore 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 pvalue, 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.
a. How large was the sample used in this survey?
We have that . We want to find n, so:
A sample of 256 was used in this survey.
b. What is the probability that the point estimate was within ±15 of the population mean?
15 is the bounds with want, 25 is the standard error. So
Z = 15/25 = 0.6 has a pvalue of 0.7257
Z = -15/25 = -0.6 has a pvalue of 0.2743
0.7257 - 0.2743 = 0.4514
45.14% probability that the point estimate was within ±15 of the population mean