Round 14.5621 to the nearest hundredth. Do not include extra zeros in your answer.
1 answer:
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Answer:
The probability will be 0.3085 or 0
Step-by-step explanation:
Given:
True mean=12.5
Sample mean =12.6
Standard deviation=0.2
Samples=100
To Find:
Probability that exceeds 12.6 ounces.
Solution:
Calculate the Z-score for given means and standard deviation.
So
Z-score= (true mean -sample mean)/standard deviation.
Z-score=(12.5 -12.6)/0.2
=-0.1/0.2
=-0.5
Now Using Z-table
P(X≥-0.5)=p(Z≥-0.5)=0.3085
Hence Probability that sample mean weight exceeds will be 0.3085
OR
By using Normal distribution with sampling ,it will be as follows
Z=(X-u)/[Standard deviation/Sqrt(No of samples)]
Z=(12.6-12.5)/(0.2/Sqrt(100)
Z=0.1/0.2/10
Z=5
So P(X≥12.6 )=P(Z≥5)=1
Pr(Z≥5)=1-1=0.
(Refer the attachment )
Hence Probability of getting ounces greater than 12.6 is '0'.
The sampling is of 0.02 size hence graphically it looks likely.
as shown in attachment.
Answer:
Step-by-step explanation:
-p(d + z) = -2z + 59
-pd - pz = -2z + 59
+pd +pd
-pz = pd - 2z + 59
+2z +2z
pz + 2z = pd + 59
(p + 2)z = pd + 59
÷(p+2) ÷(p+2)