To factor we must first find <u>factors</u> of 35
The only factors of 35 are :<em> 5 & 7 </em>
-2x is what 5 and 7 make
but only if 7 is <u>negative </u>and 5 is <u>positive </u>
so our final factors are
(x - 7) (x + 5)
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For the second equation, were going to multiple the coefficient of ( which is 2 ) by 6
We get 12
What factors of 12 add up to 7?
<em>3 and 4 </em>
So our final factors are
(2x+3)(x+2)
Answer:
a) 0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51
b) 0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51
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:
(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 17 pins is at least 51?
Here
This probability is 1 subtracted by the pvalue of Z when X = 51. So
By the Central Limit Theorem
has a pvalue of 0.9890
1 - 0.989 = 0.011
0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51
(b) What is the (approximate) probability that the sample mean hardness for a random sample of 45 pins is at least 51?
Here
has a pvalue of 0.9999
1 - 0.9999 = 0.0001
0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51