T2
spring 2014 — test No.2
2nd test – solutions
1. The probability of getting leukemia (for single person) is p = 0.00015. For a population of n = 30000 find,
using the Bernoulli-to-normal approximation:
a) the probability of four or more leukemia cases;
b) probability of less than three cases.
Hint: Bernoulli-to-normal approximation means: µ = np = 4.5 and σ 2 = npq ≈ np. σ = 2.12.
four or more means X > 3.5; standardization
P (X > 3.5) = P
3.5 − 4.5
> −0, 47 = 1 − Φ(−0.47) = 0.6808
Z=
2.12
less than three cases, means X < 2.5 or Z <
2.5−4.5
2.12
hence P (Z < −0.94) = Φ(−0.94) = 0.1736
2. [ 5 points ]A company pays its employees an average wage of $9.25 an hour with a standard deviation of
60 cents. If the wages are normally distributed and paid to the nearest cent
(a) what percentage of the workers receive wages between $8.75 and $9.69 an hour?
(b) the highest 5% of the employee hourly wages is greater than what amount?
We standardise – exactly as above and consult the tables:
8.75 − 9.25
9.69 − 9.25
(a) z8.75 =
= −0.83; z9.69 =
= 0.73
0.6
0.6
P (Z ∈ [−0.83, 0.73]) = Φ(0.73) − Φ(−0.83) = 0.565
(b)
1.645 = z0.95 =
x0.95 − 9.25
;
0.6
→ x0.95 = 10.23
3. [ 4 points ]The lifetimes, measured in miles, of 100 tires were recorded and the following results were
obtained:
x̄ = 38000 miles, S ∗ = 3200 miles.
What can you say about the distribution of X̄?
Construct a lower one-sided 99 percent interval for the mean lifetime.
Hint: The sample is big enough to put σ ≈ S ∗ and to use the quantiles of the standardised normal
distribution. The lower one-sided 99 percent interval has a 99-percent chance to cover the true µ from the
+∞ to the left.
the left boundary of the interval is
σ
S∗
3200
x̄ − √ × z1−α ≈ √ × z0.99 = 38000 − √
2.327 ≈ 37 255
n
n
100
4. Suppose that the time (measured in minutes) to repair a component is normally distributed with µ = 65
and σ = 10.
What is the probability that a component is repaired in less than one hour? [ 3 points ]
P (X < 60) = P
X − 65
60 − 65
<
10
10
= P (Z < −0.5) = 0.3085.
What is the distribution (what are µT and σT equal to) of the total time T required to repair eight
components?
T = X1 + X2 . . . + X8
T2
spring 2014 — test No.2
What is the probability that eight components are repaired in an 8-hour working day, i.e. 480 minutes?[ 4 points ]
follows a normal distribution with µT = 8×65 = 520 and variance σ 2 T = 8×102 = 800
480 − 520
T − 520
√
< √
= P (Z < −1.414) = 0.0793.
P (T < 480) = P
800
800
T = X1 +X2 . . .+X8
5. [ 4 points ]Eight measurements made on the viscosity of a motor oil blend yielded the values: x̄ =
10.23, S ∗ 2 = 0.64. Construct a 90 percent two-sided interval for µ.
Hint: The sample is NOT big enough to use the quantiles of the standardised normal distribution. We
must use the appropriate quantile of the Student’s distribution. The values ot this quantile can be found
using the TINV excel function or you may ask me to supply the adequate value.
n − 1 = ν = 7, α/2 = 0.05, t(ν = 7, 1 − α/2) = 1.895, S ∗ = 0.8,
so
√
0.8
X̄ ± t(ν = 7, 1 − α/2) × S ∗ / n = 10.23 ± 1.895 × √ = 10.23 ± 0.54.
8
6. [ 5 points ] A player checks whether the die is fair by tossing it 240 times and listing the frequencies:
1
20
40
2
46
40
3
35
40
4
45
40
5
42
40
6
52
40
Fill the empty cells of the table with the theoretical frequencies for a fair die. Using the chi-square test
verify, with α = 0.05 the hypothesis about the die being fair.
Hint: The critical region is the 95% quantile of the chi-square distribution for ν = n − 1 = 5 degrees of
freedom, which is equal to 11,07.
χ2 = (20−40)2 /40+(46−40)2 /40+(35−40)2 /40+(45−40)2 /40+(42−40)2 /40+(52−40)2 /40 = . . . 15.85.
It’s greater than the critical value (cf. above) – we must reject H0.
7. [ 5 points ] The producer of washers claims that his product thickness is equal to 1.4 mm. A sample of
n = 10 gave the results: 1.6, 1.4, 1.6, 1.5, 1.9, 1.5, 1.3, 1.4, 1.6, 1.2. Using the significance level α = 0.05
verify the hypothesis µ0 = 1.4.
Hint: The critical (rejection) region depends on your alternative hypothesis H1. From the data it seems
that H1: µ0 > 1.4 would be a more logical choice. Since the sample is small and we need to estimate
σ ≈ S ∗ a Student’s quantile will be needed.
Solution: we calculate X̄ = 1.5 and S ∗ = 0.194. The standardization (under H0) is
Z=
X̄ − 1.4 √
10 = . . . 1.63.
S∗
This value does not enter the critical region [1.833, ∞) (1.833 is the t0.95 quantile – the t value that leaves
95% of the total area to the left.