1. Using the standard normal probability table, find the following probabilities:
a) P (Z > -1.50)
P= 0.0668
b) P (-1.85 < X < 1.85) , given X is N (0,1)
P(-1.85-0 <X < 1.85-0
1
1
=P(0.9678 <X < 0.032
=0.9678 – 0.032
= 0.9356
c) P (X > 9), given that X is N (8, 4)
P= Z > (9-8)
2
P= Z> 0.5
= 0.3085
2. From the past experience it is felt that the random variable, X, the age of a mother at the
birth of her first child, is normally distributed with a mean of 26 and variance of 4 worldwide. Find the probability that a randomly selected mother:
X = 26 , N=4 , (2)
a) Has her first child before 21
P (X < 21)
= P (Z < 21-26)
2
=P (Z < -2.5)
= 0.0062
b) Has her first child after 32
P (X>32)
=P (Z > 32-26)
2
=P (Z > 3)
=0.0013
c) Has her first baby before 31
P (X < 31)
=P (Z < 31-26)
2
=P (Z < 2.5)
= 0.9938
d) Has her first child between 21 and 32
P (21 < X < 32)
=P (21 -26 < Z < 32-26)
2
2
=P (-2.5 < Z < 3)
=0.9938 – 0.0013
= 0.9925
e) Has her first child between 24 and 27
P (24 < X < 27)
=P (24 -26 < Z < 27-26)
2
2
=P (-1 < Z < 0.5)
=0.8413 – 0.3085
= 0.5328
3. A study reported in a local Malaysian newspaper that teenagers spent 8.5 hours on
average per week on the internet. The study thinks that, currently, the mean is higher.
Twenty randomly chosen teenagers were asked on how many hours per week they spend
on the internet. The sample mean was 9.0 hours with a sample standard deviation of 2.0.
Conduct a hypothesis test. The null and alternative hypotheses are?
Ho : There is the average spent time is not more than 8.5 hours.
H0: μ = 8.5,
Ha : There is the average spent time is more than 8.5 hours.
Ha: μ > 8.5
0.95
Z= 1.645
Z
= x -𝜇
𝜎
√𝑛
Z = 9.0 – 8.5
2.0
√20
Z = 0.5
= 1.118
4.4721
1.118 < 1.645
= Fail to reject Ho
0.05
4. When a new covid-19 drug is created, the pharmaceutical company must subject it to
testing before receiving the necessary permission from the Food and Drug Administration
(FDA) to market the drug world-wide. Suppose the null hypothesis is “the covid-19 drug
is unsafe.” What is the Type II Error?
Type II Error: Not to conclude the drug is safe when, in fact, it is safe.
5. Panasonic claims that its deluxe lighting bulb averages at least 20,000 hours before it
needs to be replaced. From past studies of this bulb, the standard deviation is known to be
2,000. A survey of owners of that bulb design is conducted. From the 30 lighting bulbs
surveyed, the mean lifespan was 18,500 hours with a standard deviation of 3,800 hours.
Using α=0.05, is the data highly inconsistent with the claim? (Use the steps in hypothesis
testing)
a. H0:μ≥20,000
b. Ha:μ<20,000
c. Let X =X= 18,500 hours, the average lifespan of a Panasonic bulb
d. normal distribution
e.
Z
= x -𝜇
𝜎
√𝑛
Z = 18500-20000
2000
√30
Z = −4.107 < 1.645
f. p-value=0.00004
g.
i.
ii.
alpha: 0.05
Decision: Reject the null hypothesis.
iii.
iv.
Reason for decision: The p-value is less than 0.05.
Conclusion: There is sufficient evidence to conclude that the mean
lifespan of the bulb is less than 20,000 hours.