Question

A ranch in "Smart Town" claimed that the cows they raise are smarter than the rest of the population of US cows. To prove that they announced that the average weight of their "Smart Cow" brain is 485 grams instead of the regular 458 grams. Assume that the standard deviation of Smart Cow brain weights is the same as the entire population's standard deviation=64 g.
a) What is the Probability that the Brain of a randomly selected Smart Cow will weigh at least 500 grams
b) What is the probability that the average brain weight of a sample of 36 Smart Cows will be at least 480 grams

Answer 1

Answer:

(a) The probability that the Brain of a randomly selected Smart Cow will weigh at least 500 grams is 0.4091.

(b) The probability that the average brain weight of a sample of 36 Smart Cows will be at least 480 grams is 0.6808.

Step-by-step explanation:

We are given that the average weight of their "Smart Cow" brain is 485 grams instead of the regular 458 grams.

Assume that the standard deviation of Smart Cow brain weights is the same as the entire population's standard deviation = 64 g.

Let X = weight of the brain of a randomly selected Smart Cow

So, X ~ Normal($\mu=485, \sigma^{2} = 64^{2}$)

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = population mean weight = 485 grams

            $\sigma$ = standard deviation = 64 grams

(a) The probability that the Brain of a randomly selected Smart Cow will weigh at least 500 grams is given by = P(X $\geq$ 500 grams)

        P(X $\geq$ 500 g) = P( $\frac{X-\mu}{\sigma}$ $\geq$ $\frac{500-485}{64}$ ) = P(Z $\geq$ 0.23) = 1 - P(Z < 0.23)

                                                               = 1 - 0.59095 = 0.4091

The above probability is calculated by looking at the value of x = 0.23 in the z table which has an area of 0.59095.

(b) Let $\bar X$ = sample mean weight of the brain of a randomly selected Smart Cow

The z-score probability distribution for the sample meanis given by;

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

where, $\mu$ = population mean weight = 485 grams

            $\sigma$ = standard deviation = 64 grams

            n = sample of cows = 36

Now, the probability that the average brain weight of a sample of 36 Smart Cows will be at least 480 grams is given by = P($\bar X$ $\geq$ 480 grams)

        P($\bar X$ $\geq$ 480 g) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{480-485}{\frac{64}{\sqrt{36} } }$ ) = P(Z $\geq$ -0.47) = P(Z < 0.47)

                                                               = 0.6808

The above probability is calculated by looking at the value of x = 0.47 in the z table which has an area of 0.6808.