A manufacturing process produces bags of chips whose weight is N(16 oz, 1.5 oz). On a given day, the quality control officer takes a sample of 36 bags and computed the mean weight of these bags. The probability that the sample mean weight is below 16 oz is

Group of answer choices:
A. 0.55
B. 0.45
C. 0.50
D. 0.60

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable who represent the weight for bags of chips of a population, and for this case we know the distribution for X is given by:

n=16 represent the sample size

[tex]\mu =16[/tex] represent the true mean

[tex]\sigma=1.5[/tex] represent the population standard deviation

[tex]X \sim N(16,1.5)[/tex]

Where [tex]\mu=16[/tex] and [tex]\sigma=1.5[/tex]

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

The probability that the sample mean weight is below 16 oz is:

[tex]P(\bar x<16)=P(Z<\frac{16-16}{\frac{1.5}{\sqrt{36}}})=P(Z<0)=0.5[/tex]

So the best option for this case is 0.5