Find the seventh term of the following sequence?
1 answer:
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Answer:
a) 0.6915; b) 2.83 years
Step-by-step explanation:
For part a,
The formula for a z score is
Our mean, μ, is 3.25 and our standard deviation, σ, is 0.5.
This gives us
z = (3-3.25)/0.5 = -0.25/0.5 = -0.5
Using a z table, we see that the area under the curve to the left of this is 0.3085. However we want the area to the right; this means we subtract from 1:
1-0.3085 = 0.6915
For part b,
We look in the cells of a z table to find the value closest to 20%, or 0.2000. This is 0.2005, which corresponds with a z score of -0.84:
-0.84 = (X-3.25)/0.5
Multiply both sides by 0.5:
0.5(-0.84) = ((X-3.25)/0.5)(0.5)
-0.42 = X-3.25
Add 3.25 to each side:
-0.42+3.25 = X-3.25+3.25
X = 2.83
The advertised life would be 2.83 years.
Answer:
0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Also, important to remember that the standard deviation is the square root of the variance.
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
, the sample means with size n of at least 30 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 be greater than 70.5 kilograms?
This is 1 subtracted by the pvalue of Z when X = 70.5. So
By the Central limit theorem
has a pvalue of 0.7764
1 - 0.7764 = 0.2236
0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.