Answer:
P(10 ≤ x ≤ 12) = 0.4274
Step-by-step explanation:
Population mean = u = 10
Population Standard Deviation = [tex]\sigma[/tex] = 9
Sample size = n = 43
Sample mean([tex]\mu_{x}[/tex]) is equal to the population mean. So,
Sample mean = [tex]\mu_{x}[/tex] = 10
Sample standard deviation([tex]\sigma_{x}[/tex]) is equal to population standard deviation divided by square root of sample size. So,
Sample standard deviation = [tex]\sigma_{x}[/tex] = [tex]\frac{9}{\sqrt{43}}=1.372[/tex]
We have to find the probability that for a random sample of n = 43, the value lies between 10 and 12 i.e. P(10 ≤ x ≤ 12)
P(10 ≤ x ≤ 12) = P(x ≤ 12) - P( x ≤ 10)
We can find P(x ≤ 12 ) and P(x ≤ 10) by converting these values to z scores.
The formula for z score is:
[tex]z=\frac{x-\mu_{x}}{\sigma_{x}}[/tex]
For x =12, we get:
[tex]z=\frac{12-10}{1.3725}=1.457[/tex]
For x =10, we get:
[tex]z=\frac{10-10}{1.3725}=0[/tex]
So,
P(x ≤ 12) - P( x ≤ 10) = P(z ≤ 1.457) - P(z ≤ 0)
From the z table,
P(z ≤ 1.457) = 0.9274
P(z ≤ 0) = 0.5
So,
P(x ≤ 12) - P( x ≤ 10) = P(z ≤ 1.458) - P(z ≤ 0) = 0.9274 - 0.5 = 0.4274
So,
P(10 ≤ x ≤ 12) = P(x ≤ 12) - P( x ≤ 10) = 0.4274
Therefore,
The probability that for a random sample of size 43, the mean lies between 10 and 12 is 0.4274.
The probability that P(10 ≤ x ≤ 12). is 42.79%.
The z score is used to determine by how many standard deviations, the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\where\ x=raw\ score,\mu=mean, \sigma=standard\ deviation,n= sample\ size\\\\\\Given\ that\ \mu=10, \sigma=9,n=43\ hence:\\\\For\ x=10:\\\\z=\frac{10-10}{9/\sqrt{43} }=0\\\\\\ For\ x=12:\\\\z=\frac{12-10}{9/\sqrt{43} }=1.46\\\\\\[/tex]
From the normal distribution table, P(10 ≤ x ≤ 12) = P(0 ≤ z ≤ 1.46) = P(z < 1.46) - P(z<0) = 0.9279 - 0.5 = 42.79%.
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