Answer:
ai ) [tex]P(\= X \le 3.0 ) =0.980[/tex]
aii) [tex]P(2.65 \le \= X \le 3.00) = 0.480[/tex]
b ) [tex]n = 32[/tex]
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 2.65[/tex]
The standard deviation is [tex]\sigma = 0.85[/tex]
Let the random sediment density be X
given that the sediment density is normally distributed it implies that
X N(2.65 , 0.85)
Now probability that the sample average is at 3.0 is mathematically represented as
[tex]P(\= X \le 3.0 ) = P[\frac{\= X - \mu} {\frac{\sigma }{\sqrt{n} } } \le \frac{3.0 - \mu}{\frac{ \sigma}{\sqrt{n} } } ][/tex]
Here n is the sample size = 25 and [tex]\= X[/tex] is the sample mean
Now Generally the Z-value is obtained using this formula
[tex]Z = \frac{\= X - \mu} {\frac{\sigma }{\sqrt{n} } }[/tex]
Thus
[tex]P(\= X \le 3.0 ) = P[Z \le \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{25} } } ][/tex]
[tex]P(\= X \le 3.0 ) = P[Z \le 2.06 ][/tex]
From the z-table the z-score is 0.980
Thus
[tex]P(\= X \le 3.0 ) =0.980[/tex]
Now probability that the sample average is between 2.65 and 3.00 is mathematically evaluated as
[tex]P(2.65 \le \= X \le 3.00) = P [\frac{2.65 - \mu }{ \frac{\sigma }{\sqrt{n} } } < \frac{\= X - \mu }{ \frac{\sigma }{\sqrt{n} } } < \frac{3.0 - \mu }{ \frac{\sigma }{\sqrt{n} } }][/tex]
[tex]P(2.65 \le \= X \le 3.00) = P [\frac{2.65 - 2.65 }{ \frac{0.85 }{\sqrt{25} } } <Z < \frac{3.0 - 2.65 }{ \frac{0.85 }{\sqrt{25} } }][/tex]
[tex]P(2.65 \le \= X \le 3.00) = P[0 < Z< 2.06][/tex]
[tex]P(2.65 \le \= X \le 3.00) = P(Z < 2.06) - P(Z<0)[/tex]
From the z-table
[tex]P(2.65 \le \= X \le 3.00) = 0.980 - 0.50[/tex]
[tex]P(2.65 \le \= X \le 3.00) = 0.480[/tex]
Now from the question
[tex]P(\= X \le 3.0 ) =0.99[/tex]
=> [tex]P(\= X \le 3.0 ) = P[Z \le \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } } ] = 0.99[/tex]
Generally the critical value of z for a one tail test such as the one we are treating that is under the area 0.99 is [tex]t_z = 2.33[/tex] this is obtained from the critical value table
So
[tex]t_z = \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } }[/tex]
[tex]2.33 = \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } }[/tex]
=> [tex]n = 32[/tex]
