Question

Let x be a random variable that represents the weights in kilograms (kg) of healthy adult female deer (does) in December in a national park. Then x has a distribution that is approximately normal with mean ? = 64.0 kg and standard deviation ? = 8.3 kg. Suppose a doe that weighs less than 55 kg is considered undernourished.

(a) What is the probability that a single doe captured (weighed and released) at random in December is undernourished? (Round your answer to four decimal places.)


(b) If the park has about 2200 does, what number do you expect to be undernourished in December? (Round your answer to the nearest whole number.)
does

(c) To estimate the health of the December doe population, park rangers use the rule that the average weight of n = 60 does should be more than 61 kg. If the average weight is less than 61 kg, it is thought that the entire population of does might be undernourished. What is the probability that the average weight  x for a random sample of 60 does is less than 61 kg (assuming a healthy population)? (Round your answer to four decimal places.)

(d) Compute the probability that  x < 65 kg for 60 does (assume a healthy population). (Round your answer to four decimal places.)

Suppose park rangers captured, weighed, and released 60 does in December, and the average weight was  x = 65 kg. Do you think the doe population is undernourished or not? Explain.

Since the sample average is below the mean, it is quite unlikely that the doe population is undernourished.

Since the sample average is above the mean, it is quite unlikely that the doe population is undernourished.    

Since the sample average is above the mean, it is quite likely that the doe population is undernourished.

Since the sample average is below the mean, it is quite likely that the doe population is undernourished.

Answer 1

Answer a) 0.1401

Step 3: Use the standard normal table to conclude that:

P (Z<?1.08) = 0.1401 (Screenshot attached)

Answer b) 308

Expected number undernourished does in December = Total Number of Does * Probablity that doe is undernourished

Expected number undernourished does in December = 2200*0.1401

Expected number undernourished does in December = 308.22

Expected number undernourished does in December = 308 (Rounded to nearest whole number)

Answer c) 0.3594

Step 3: Use the standard normal table to conclude that:

P (Z<?0.36) = 0.3594 (Screenshot attached)

Answer d) 0.5478

Step 3: Use the standard normal table to conclude that:

P (Z<0.12) = 0.5478 (Screenshot attached)

Step 1: Set up null and alternative hypotheses.

H₀: ? = 61
H₁: ? > 61  

Step 2: Determine ? (level of significance of hypothesis test).

? = 0.05 (Note: ? = level of significance of hypothesis test = probability of making Type I error.)

Step 3: Calculate Test Statistic using x?, n, and ?

x? (sample mean) =  
n (sample size) =  
? (population standard deviation) = 8.3

Test Statistic = (x? - ?)/^?/_?n

Test Statistic = (65-61)/8.3/?60

Test Statistic = 3.733

Step 4: Find Critical Value

Critical Value = z_? = z-score corresponding to the right-tailed area.

Critical Value = z_? = 1.645 (Obtained using z distribution table)

Step 5: Make Decision
Test Statistic = 3.733

Critical Value is 1.64
In this case, Test Statistic is greater than or equal to critical value.
If the test statistic is greater than or equal to the critical value, then reject the null hypothesis H_o: ? = 61.

Conclusion: This means there is sufficient evidence that the doe population is not undernourished.

Since the sample average is above the mean, it is quite unlikely that the doe population is undernourished.