Answer:
0.9545
Step-by-step explanation:
Given that :
Mean = 300
Standard deviation, σ = 18
Sample size, n = 144
Zscore = (x - mean) ÷ σ/√n
At score, x = 297
Zscore = (297 - 300) / (18/12) = - 2
P(Z< - 2) = 0.02275
At score, x = 303
Zscore = (303 - 300) / (18/12) = 2
P(Z< 2) = 0.97725
P(Z < 2) - P(Z < - 2) = 0.97725 - 0.02275 = 0.9545
The probability that the sample mean will be between 297 to 303 is 0.9545.
We have given that ,A population has a mean of 300 and a standard deviation of 18. A sample of 144 observations will be taken.
Mean = 300
Standard deviation, σ = 18
Sample size, n = 144
We have to find the probability that the sample mean will be between 297 to 303
Therefore we use the formula of Z score.
[tex]Zscore = \frac{ (x - mean)}{\sigma/\sqrt n}[/tex]
At the score, x = 297
Zscore = (297 - 300) / (18/12) = - 2
P(Z< - 2) = 0.02275
At score, x = 303
Zscore = (303 - 300) / (18/12) = 2
P(Z< 2) = 0.97725
Therefore we get,
P(Z < 2) - P(Z < - 2) = 0.97725 - 0.02275 = 0.9545
Therefore,the probability that the sample mean will be between 297 to 303 is 0.9545.
To learn more about the probability visit:
https://brainly.com/question/25870256
