kamari opens a bank account with 250 five weeks her account balances 800 what was the average rate of change of her account bala
1 answer:
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Answer:
There is a 0.82% probability that a line width is greater than 0.62 micrometer.
Step-by-step explanation:
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
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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.
In this problem
The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so .
What is the probability that a line width is greater than 0.62 micrometer?
That is
So
Z = 2.4 has a pvalue of 0.99180.
This means that P(X \leq 0.62) = 0.99180.
We also have that
There is a 0.82% probability that a line width is greater than 0.62 micrometer.
Answer:
hello attached below is the required table that is missing and the completed table as well
A) mean deviation = 0.1645
B) 2/3
Step-by-step explanation:
From the table we calculated the midpoint value (x) and c.f
N = 27
median class = 2.35 to 2.45
median = l + [ (N/2) - cf / F ] * H
= 2.35 + [ (13.5 - 13) / 6 ] *0.1 = 2.3583
hence mean deviation by median
= summation of fi |xi -M| / summation of fi
= 4.4417 / 27 = 0.1645
B ) probability of getting an odd number or prime number or both
the probability of an odd number = 1/2
also the probability of an even number = 1/2
while the probability of getting neither of them = 1/3
hence the probability of getting odd or prime or both
= 1/2 + 1/2 - 1/3 = 2/3
