Write an expression for the sequence of operations described below.
1 answer:
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Let , side lawn is a.
Area of boundary is :
Now, total cost is given by :
Area left is :
Price to empty space with turfs is :
Therefore, the cost of covering the empty space with turfs at the rate of Rs 20 per sq. metre is Rs 262500.
Hence, this is the required solution.
Answer:
0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months
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 can be 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
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
In this problem, we have that:
What is the probability that the sample mean would differ from the true mean by less than 1.11 months?
This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So
X = 120.1
By the Central Limit Theorem
has a pvalue of 0.7517
X = 117.9
has a pvalue of 0.2483
0.7517 - 0.2483 = 0.5034
0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months