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 μ = 60.0 kg and standard deviation σ = 8.4 kg. Suppose a doe that weighs less than 51 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 2350 does, what number do you expect to be undernourished in December? (Round your answer to the nearest whole number.)

(c) To estimate the health of the December doe population, park rangers use the rule that the average weight of n = 50 does should be more than 57 kg. If the average weight is less than 57 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 50 does is less than 57 kg (assuming a healthy population)? (Round your answer to four decimal places.)

(d) Compute the probability that x < 61.2 kg for 50 does (assume a healthy population). (Round your answer to four decimal places.)

Answer 1

Solution :

mean = = 60.0

standard deviation = = 8.4

a) P(x < 51) = P[(x - ) / < (51 - 60) /8.4 ]

= P(z < -1.07)

= 0.1423

probability =0.1423

b)

P(x < 51) = P[(x - ) / < (51 - 60) /8.4 ]

= P(z < -1.07)

= 0.1423

probability =0.1423 * 2350 = 334

Answer =334

c)

n = 50

= = 60.0

= / n = 8.4/ 50 = 1.1879

P( > 57) = 1 - P( < 57)

= 1 - P[( - ) / < (57 -60) / 1.1879]

= 1 - P(z < -2.53 )

= 1 - 0.0057 = 0.9943

Probability = 0.9943

P( < 57) = P(( - ) / < (57 -60) /1.1879 )

= P(z < -2.53 )

= 0.0057

probability = 0.0057

d)

P( < 61.2 ) = P(( - ) / < (61.2 -60) /1.1879 )

= P(z < 1.01 )

= 0.8438

probability = 0.8438