Question

Suppose the bacteria count per milliliter of raw milk has mean 2500 and standard deviation 300. If 46 samples of raw milk are randomly chosen, what is the probability that the mean bacteria count will be greater than 2620?

Answer 1

Probability that the mean bacteria count will be greater than 2620 is 0.0034.

Step-by-step explanation:

We are given that the bacteria count per milliliter of raw milk has mean 2500 and standard deviation 300. 46 samples of raw milk are randomly chosen.

Assuming data follows normal distribution.

Let $\bar X$ = sample mean bacteria count

The z-score probability distribution for sample mean is given by;

                       Z  = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where,  $\mu$ = mean bacteria count per milliliter of raw milk = 2500

            $\sigma$ = standard deviation = 300

            n = sample of raw milk = 46

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, probability that the mean bacteria count will be greater than 2620 is given by = P($\bar X$ > 2620)

     P($\bar X$ > 2620) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{2620-2500}{\frac{300}{\sqrt{46} } }$ ) = P(Z > 2.71) = 1 - P(Z $\leq$ 2.71)

                                                             = 1 - 0.9966 = 0.0034

The above probability is calculated by looking at the value of x = 2.71 in the z table which gives an area of 0.9966.

Hence, the probability that the mean bacteria count will be greater than 2620 is 0.0034.