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What is a31 of the arithmetic sequence for which a5=12.4 and a9=22.4
1 answer:
For the arithmetic sequence
a₁, a₂, a₃, ...,
the n-th term is
where d = the common difference
Because a₅ = 12.4,
a₁ + 4d = 12.4 (1)
Because a₉ = 22.4,
a₁ + 8d = 22.4 (2)
Subtract (1) from (2).
a₁ + 8d - (a₁ + 4d) = 22.4 - 12.4
4d = 10
d = 2.5
From (1),
a₁ = 12.4 - 4*2.5 = 2.4
Therefore
a₃₁ = 2.4 + 30*2.5 = 77.4
Answer: a₃₁ = 77.4
a₁, a₂, a₃, ...,
the n-th term is
where d = the common difference
Because a₅ = 12.4,
a₁ + 4d = 12.4 (1)
Because a₉ = 22.4,
a₁ + 8d = 22.4 (2)
Subtract (1) from (2).
a₁ + 8d - (a₁ + 4d) = 22.4 - 12.4
4d = 10
d = 2.5
From (1),
a₁ = 12.4 - 4*2.5 = 2.4
Therefore
a₃₁ = 2.4 + 30*2.5 = 77.4
Answer: a₃₁ = 77.4
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Answer:
15.15% probability that the sample mean will be $192,000 or more.
Step-by-step explanation:
To solve this problem, 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 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:
The probability that the sample mean will be $192,000 or more is
This is 1 subtracted by the pvalue of z when X = 192000. So
By the Central Limit Theorem
has a pvalue of 0.8485.
1-0.8485 = 0.1515
15.15% probability that the sample mean will be $192,000 or more.