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 u 70.0 kg and standard deviation o 7.3 kg. Suppose a doe that weighs less than 61 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 2600 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 n55 does should be more than 67 kg. If the average weight is less than 67 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 55 does is less than 67 kg (assuming a healthy population)? (Round your answer to four decimal places.) (d) Compute the probability that x< 71.5 kg for 55 does (assume a healthy population). (Round your answer to four decimal places.) Suppose park rangers captured, weighed, and released 55 does in December, and the average weight was x 71.5 kg. Do you think the doe population is undernourished or not? Explain. 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. e Since the sample average is above the mean, it is cuite unlikely that the doe population is undernourished. Since the sample average is below the mean, it is quite unlikely that the doe population is undernourished.

Answer 1

Pg. No:1 Solution Given that → Let x be a srandom vaciable weights Pn kilograme → mean as 7000 kg → standaed deviation a=7.3kg. >> weight less than 61 kg The probability that a single does captured at random → weight ley than blkg . P(x <61) aprelu 5-4). p(za 61-70 G1-70 7.3 It fond =plze-19398) from pualue fom 0:109 P(x261) = 0.109 | 10 pack has . 2000 . pook & P(x261

where, Pg No - ③ pack = 2600, P(x261) = 0.109 = 9600 x 0.109 = 283.14 & neareast whole number is a83 | → dverage weight n = 55 mae than u = 70 kg → standard deviation : 5:5-7:3] 1.3 55 → +(7<67) = p ( 69) 67 - 70 0.9843 09843 =P(z< Balus) Ap(Z < 3.0678) = 0.001183 Sat = 0:00183 find in pralue - PCXc67)

Paino 73 o probability 71.5kg P(82Hs) -rf < 25) = p (za 71.5-70 9843) =plze da sus) -p(221.5239) = 0.06425 ip(t < 71,5) = 0.06h35 is Since the Sample average it is quite unlikely that undeenourished. is above the mean, doe population is