Exam 2 f12
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Name__________________________________
1) Mr. Praline runs a pet shop. 35% of the time, the parrots he sells to customers turn out to be (upon close examination)
dead.
(a) Find the probability (to four decimal places) that less than 40% of 400 customers purchase a dead parrot.
𝑛 = 400
𝑝̂ < 0.40 𝑍 <
𝑝 = 0.35
0.40 − 0.35
𝑍 < 2.10
0.35∗(1−0.35)
√
400
0.9821
(b) Find the probability (to four decimal places) that less than 40% of 900 customers purchase a dead parrot.
𝑛 = 900
𝑝̂ < 0.40 𝑍 <
𝑝 = 0.35
0.40 − 0.35
𝑍 < 3.14
0.35∗(1−0.35)
√
900
0.9992
(c) Find the probability (to four decimal places) that less than 40% of 1600 customers purchase a dead parrot.
𝑛 = 1600
𝑝̂ < 0.40 𝑍 <
𝑝 = 0.35
0.40 − 0.35
√
𝑍 < 4.19
0.35∗(1−0.35)
1600
1
2. Whale-weigh-stations X and Y are used for weighing whales, but readings vary from weighing to weighing. A certain
blue whale is weighed repeatedly at each station. The weights given for it by station X vary normally with mean 82 tons
and standard deviation 5 tons. The weights given for it by station Y vary normally with mean 77 tons and standard
deviation 7 tons. (Be as accurate as possible.)
(a) Find the probability that the average of 2 readings from station X is less than 84 tons.
𝑋 = 𝑁(82, 5)
𝑋̅ = 𝑁 (82,
𝑋̅ < 84
𝑧<
84 − 82
5
( 2)
5
√2
)
𝑧 < 0.57
0.7157
√
(b) Find the probability that the average of 9 readings from station X is less than 84 tons.
𝑋̅ < 84
𝑧<
84 − 82
5
√9
( )
𝑧 < 1.20
0.8849
(c) Find the probability that the average of 30 readings from station X is less than 84 tons.
𝑋̅ < 84
𝑧<
84 − 82
5
)
√30
(
𝑧 < 2.19
0.9857
(d) Find the probability that the reading from station X is less than the reading from station Y.
𝑋 = 𝑁(82, 5)
𝑌 = 𝑁(77, 7)
𝑋 − 𝑌 = 𝑁 (82 − 77, √52 + 72 ) = 𝑁(5, 8.6023)
𝑋<𝑌
𝑋−𝑌 <0
𝑍<
0−5
8.6023
𝑍 < −0.58
0.2810
(e) Find the probability that the average of 25 readings from station X is less than the average of 25 readings from station
Y.
𝑋 = 𝑁(82, 5)
𝑋̅ = 𝑁 (82,
5
√25
𝑌 = 𝑁(77, 7)
= 1) 𝑌̅ = 𝑁 (77,
7
√25
= 1.4)
𝑋̅ − 𝑌̅ = 𝑁 (82 − 77, √12 + 1.42 ) = 𝑁(5, 1.7205)
𝑋̅ < 𝑌̅
𝑋̅ − 𝑌̅ < 0
𝑍<
0−5
1.7205
𝑍 < −2.91
0.0018
3.Mr. Wenslydale runs a cheese shop, but the probability isn’t high that he’ll actually have cheese to sell in the shop on
any given day. A simple random sample of 169 days reveals that on 12% of those days he actually has cheese to sell.
(For the confidence intervals, answers to three decimal places.) (a) A disgruntled regular customer claims that he only has
cheese 9% of the time. Do we have evidence that the customer is underestimating the truth at each of the following levels
(show work, including hypotheses, test-statistic, and p-value, then circle correct answer in each case)?
𝑝̂ = 0.12
𝑛 = 169
𝑝 = proportion of days he has cheese
𝐻0 : 𝑝 = 0.09
𝐻𝑎 : 𝑝 > 0.09
𝑝̂ ≥ 0.12
𝑍≥
0.12 − 0.09
0.09∗(1−0.09)
√
169
𝑍 ≥ 1.36
𝑝 − value = 0.0869 (right tail)
0.1%
0.0869 ≮ .001
yes
no
1%
0.0869 ≮ 0.01
yes
no
3%
0.0869 ≮ 0.03
yes
no
5%
0.0869 ≮ 0.05
yes
no
10%
0.0869 < 0.10
yes
no
15%
0.0869 < 0.15
yes
no
20%
0.0869 < 0.20
yes
(b) We can be 95% confident that the proportion of days Mr. Wenslydale has cheese is between what two numbers?
𝑝̂ (1 − 𝑝̂ )
0.12(1 − 0.12)
𝑝̂ ± 𝑧√
= 0.12 ± 1.960 ∗ √
𝑛
169
0.012 ± 0.049 (0.071, 0.169)
no
(c) Provided our sample of days isn’t among the 1 out of 1000 most extreme cases, the proportion of days Mr. Wenslydale
has cheese is between what two numbers?
𝑝̂ (1 − 𝑝̂ )
0.12 (1 − 0.12)
𝑝̂ ± 𝑧√
= 0.12 ± 3.291 ∗ √
𝑛
169
0.012 ± 0.082 (0.038, 0.202)
4.The Amazing Mystico and Janet build apartment flats by hypnosis. The time such buildings remain standing has
unknown mean  and known standard deviation 3.5 days. A simple random sample of 81 such buildings provides a mean
time standing of 20.3 days. (For the confidence intervals, answers to two decimal places.)
(a) Find a 96% confidence interval for the mean standing time of all such buildings.
𝑥̅ = 20.3
𝑥̅ ± 𝑧
𝜎
√𝑛
= 20.3 ± 2.054
3.5
√81
𝑛 = 81
𝜎 = 3.5
= 20.35 ± 0.80 gives an interval of (19.50, 21.10)
(b) Find an 80% confidence interval for the mean standing time of all such buildings.
𝑥̅ = 20.3
𝑥̅ ± 𝑧
𝜎
√𝑛
= 20.3 ± 1.282
3.5
√81
𝑛 = 81
𝜎 = 3.5
= 20.35 ± 0.50 gives an interval of (19.80, 20.80)
(c) Mystico claims that on average his buildings last 21 days. Do we have evidence at each of the following levels that
he’s exaggerating (show work, including hypotheses, test-statistic, and p-value, then circle correct answer in each case)?
𝜇 = average of all building lifetimes
𝐻0 : 𝜇 = 21
𝐻𝑎 : 𝜇 < 21
Left tail is:
𝑥̅ ≤ 20.3 𝑍 ≤
20.3 − 21
3.5
)
√81
(
𝑍 ≤ −1.80
𝑝 − value = 0.0359
0.1%
0.0359 ≮0.001
yes
no
1%
0.0359 ≮ 0.01
yes
no
3%
0.0359 ≮ 0.03
yes
no
5%
0.0359 < 0.05
yes
no
10%
0.0359 < 0.10
yes
no
15%
0.0359 < 0.15
yes
no
20%
0.0359 < 0.20
yes
no
(d) The Ministry of Housing (which doesn’t care one way or the other) claims that on average his buildings last 21 days.
Should we reject their at each of the following levels? (show work, including hypotheses, test-statistic, and p-value, then
circle correct answer in each case)?
𝐻0 : 𝜇 = 21
𝐻𝑎 : 𝜇 ≠ 21
𝑝 − value = 0.0359 ∗ 2 = 0.0718
0.1%
0.0718 ≮0.001
reject don’t
1%
0.0718 ≮0.01
reject don’t
3%
0.0718 ≮0.03
reject don’t
5%
0.0718 ≮0.05
reject don’t
10%
0.0718 < 0.10
15%
0.0718 < 0.15
20%
0.0718 < 0.20
reject don’t
reject don’t
reject don’t