Download fall/winter 2011/12 — REVIEW PROBLEMS No. 2 — 1 NORMAL

fall/winter 2011/12 — REVIEW PROBLEMS No. 2 —
1
NORMAL DISTRIBUTION, POINT AND INTERVAL ESTIMATION,
HYPOTHESES
1. The time for worker to assemble a component is normally distributed with mean µ = 15
minutes and standard deviation σ = 2.
Denote the mean assembly times of 16 day-shift workers and 9 night-shift workers by X̄
and Ȳ , respectively. Assume that the assembly times of workers are mutually independent. (1) Describe the distribution of Z = X̄ − Ȳ ; (2) Compute P (Z < −1.5).
Hint: P = 0.0359
2. A cable car has a load capacity of 5000 lb. Assume that the weight of a randomly selected
person is a normally distributed variable W with µ = 175 lb, and σ = 20 lb. Determine
the maximum number of persons allowed on the car so that their total weight exceeds the
5000 lb. weight limit with a probability less than 0.05.
Hint: n = 27.
3. The IQs of 600 applicants to a certain college are normally distributed with a mean
µ = 115 and a standard deviation σ = 12. If the college requires an IQ of at least 95,
how many of these students will be rejected (on this basis)?
Hint: Find for how many students their IQs are less that 94.5.
4. A process for manufacturing an electronic component is 0.1% defective. (On the average, one component in 1000 is not usable.) Use the normal-curve approximation to the
Bernoulli scheme to find the probability of obtaining a sample of 100 randomly selected
components without any one of them being defective.
Hint: The expected value of a Bernoulli-distributed RV is µ = np, and its variance –
σ 2 = npq. No defects” means that the number” of defective components is less than 0.5.
”
”
5. The random variable X, representing the number of cherries in a cherry pudding, has the
following probability distribution:
x
P (X = x)
4
0.2
5
0.4
6
7
0.3 0.1
(a) Find the mean µ and the variance σ 2 of X.
2
(b) Find the mean µX and the variance σX
of the mean X of 36 cherry puddings.
(c) Find the probability that the average number of cherries in 36 cherry puddings will
be less than 5.5.
√
Hint: The mean X of n cherry puddings has a normal distribution N (µ, σ/ n). The µ
and σ values are these calculated in (a).
6. A soft-drink machine is regulated so that the amount of drink dispensed is approximately
normally distributed with a standard deviation of 0.15 deciliter. Find a 95% confidence
fall/winter 2011/12 — REVIEW PROBLEMS No. 2 —
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interval for the mean of all drinks dispensed by this machine if a random sample of 36
drinks has an average content of 2.25 deciliters.
Hint: Standard procedure. The interval is to be two-sided (symmetric), so you must add
to and subtract from
√ 2.25 the appropriate quantile of the normal distribution, u(1 − α/2)
multiplied by σ/ n.
7. A random sample of shearing pins are taken in a study of the Rockwell hardness of the
head on the pin. Measurements on the Rockwell hardness were made for each of the 12,
yielding an average value (X̄) of 48.50, and the sample unbiased estimator (S ∗ ) of 1.5.
Construct a 90% confidence interval for the mean Rockwell hardness.
Hint: As above – standard procedure. The interval is to be two-sided (symmetric),
so you must add to and subtract from 48.50
√ the appropriate quantile of the Student’s
distribution, t(1 − α/2, ν) multiplied by S ∗ / n.
8. A manufacturer of sports equipment has developed a new synthetic fishing line that he
claims has a mean breaking strength of 8 kilograms with a standard deviation of 0.5
kilogram. Test the H0 hypothesis that µ = mu0 = 8 against the alternative that µ 6= 8
if a random sample of n = 50 lines is tested and found to have a mean breaking strength
of x̄ = 7.8 kilograms. Use α = 0.01
Solution: The u(1 − α/2) quantile of the standardized normal distribution is equal to
2.575; the standardized mean
x̄ − µ0
√ =?
σ/ n
The hypothesis has to rejected.