Nadia charges $7.00 an hour for babysitting. She babysits 19.50 hours the first week of the month and 19 hours the second week o
f the month. Suppose Nadia raises her rate by $0.70 an hour. How many hours would she need to work to earn the same amount of money she made in the first two weeks of the month?
1 answer:
You might be interested in
Answer:
52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem:
Normal probability distribution:
Problems of normally distributed samples are solved using 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 random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct
This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 - 339 = 30054. So
X = 30732
By the Central Limit Theorem
has a pvalue of 0.7642.
X = 30054
has a pvalue of 0.2358
0.7642 - 0.2358 = 0.5284
52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct