Length = 20 ft
If he uses 60 ft of fencing to line his rectangular backyard, then the perimeter is 60 ft.
Perimeter formula is ((Width × 2) + (Length × 2))
10 × 2 = 20
20 × 2 = 40
40 + 20 = 60
Luke invested $2400 in two accounts. The first account pays 3% simple interest. The second account pays 3% compound interest wit
1 answer:
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Answer:
54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches
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 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 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.
In this problem, we have that:
What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches
Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.
Lesser than 207.9.
pvalue of Z when X = 207.9. So
By the Central Limit Theorem
has a pvalue of 0.2743
2*0.2743 = 0.5486
54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches