Solutions of the equation: z2 - 12z + 36 = 0?
1 answer:
36 becomes a negative
First -36 + -1 -37
Next: -18 + -2 = -20
Next: -12 + -3 = -15
Next: -9 + -4 which equals -13
Finally we get to the last part: -6 + -6 = -12
We change z2 to
- 6(z) = 36
z(z) - 6
Because, we add up for the first two terms
Switch up, up the problem 6(z) - 6
Now, let's add the number 4 on the terms
z - 6 (z - 6)
Then, let's multiply z - 6 from z - 6
From this problem we have to add 6 from the sides that we are working with
Therefore, your answer for z is 6
First -36 + -1 -37
Next: -18 + -2 = -20
Next: -12 + -3 = -15
Next: -9 + -4 which equals -13
Finally we get to the last part: -6 + -6 = -12
We change z2 to
z(z) - 6
Because, we add up for the first two terms
Switch up, up the problem 6(z) - 6
Now, let's add the number 4 on the terms
z - 6 (z - 6)
Then, let's multiply z - 6 from z - 6
From this problem we have to add 6 from the sides that we are working with
Therefore, your answer for z is 6
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Answer:
a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.
b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.
c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2
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 normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, 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.
The mean of a population is 74 and the standard deviation is 15.
This means that
Question a:
Sample of 36 means that
This probability is 1 subtracted by the pvalue of Z when X = 78. So
By the Central Limit Theorem
has a pvalue of 0.9452
1 - 0.9452 = 0.0548
0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.
Question b:
Sample of 150 means that
This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So
X = 77
has a pvalue of 0.9929
X = 71
has a pvalue of 0.0071
0.9929 - 0.0071 = 0.9858
0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.
c. A random sample of size 219 yielding a sample mean of less than 74.2
Sample size of 219 means that
This probability is the pvalue of Z when X = 74.2. So
has a pvalue of 0.5793
0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2