Answer:
0.0021
Step-by-step explanation:
Mean weight = u = 6.07 ounce
Standard deviation = [tex]\sigma[/tex] = 0.1 ounce
Sample size = n = 33
Test value = x = 6.02
We need to find the probability that mean weight of the sample is less than 6.0 ounces. Mathematically this can be expressed as:
P(x < 6.02)
In order to find this probability we need to convert this value to z-score. The formula to calculate the z score for the sample is:
[tex]z=\frac{x-u}{\frac{\sigma}{\sqrt{n}}}[/tex]
Using the values, we get:
[tex]z=\frac{6.02-6.07}{\frac{0.1}{\sqrt{33}}}\\\\ z = -2.87[/tex]
Thus, the probability that mean weight of the sample is less than 6.0 ounces is equivalent to P( z < - 2.87 )
i.e.
P(x < 6.02) = P( z < - 2.87 )
From the z table, the probability of z score being less than -2.87 comes out to be 0.0021
So,
P(x < 6.02) = 0.0021
Therefore, the probability that the mean weight of the sample is less than 6.02 ounces is 0.0021
