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STT 315, Section 201 Worksheet 4_2 7/28/2014 SOLUTIONS
1. The candy company claims that 10% of the M&M’s it produces are green. Suppose that the
candies are packaged at random in small bags containing about 50 M&M’s. A class of
elementary school students learning about percent open several bags, counts the various colors of
the candies and calculates the proportion that are green.
a) Can we approximate this to a normal model? No, n*p=0.1*0=5<9
b) How many M&M’s do we require in a bag to be able to approximate a normal model?
n*0.1>9 => n>90, so should be at least 91
2. It is generally believed that near sightedness affects about 12% of all children. A school
district has registered 170 incoming kindergarten children.
a) Can you approximate a normal model? Yes, n*p=0.12*170=20.4
b) How many of the incoming students might the school expect to be near sighted?
Approximately 20 students are expected to be near sighted.. x=n*p=20.4
3. Based on past experience, a bank believes that 7% of the people who receive loans will not
make payments on time. The bank has recently approved 200 loans.
a) What are the mean and standard deviation of the proportion of clients in this group who may
not make timely payments? Mean=p=0.07 SD(p)=√p*(1-p)/n=√ =0.018
b) Can a normal model be approximated? Yes, n*p and n*(1-p) >9
c) What’s the probability that over 10% of these clients will not make the payment?
P(p>0.1)=Normalcdf(0.1,9^99,0.07,0.018)=0.0478
d) What is the probability that 3% to 9% of these clients will not make a payment?
P(0.03<p<0.09)=Normalcdf(0.03,0.09,0.07,0.018)=0.8536
e) What proportion of people will have at least a 60% probability of not making a payment?
P0=Invnorm(0.6,0.07,0.018)=0.0746
4. A restaurateur anticipates serving about 180 people on a Friday evening and believes that 20%
of all the customers will order the chef’s steak special. If he desires to satisfy all the customers at
least 97% of the time, how many steaks must he make?
STT 315, Section 201 Worksheet 4_2 7/28/2014 To calculate the proportion of steaks such that 97% of the customers are satisfied is p̂
=InvNorm(0.97,0.2,√(0.2*0.8/180))=0.256, So the number of steaks that the restaurateur
that the chef would make is 180*0.256=46.08, approx. 46 to 47 steaks.
5. Assume that the duration of human pregnancies can be described as a normal model with
mean 266 days and standard deviation 16 days.
a) What percentage of pregnancies should last between 279 and 280 days?
Normalcdf(279,280,266,16)=0.0174
b) At least how many days should the longest 25% of all pregnancies last?
InvNorm(0.75,266,16)=276.79
c) Suppose a certain obstetrician is currently providing prenatal care to 60 pregnant women. Let
y (bar) represent the mean length of their pregnancies. According to CLT what is the distribution
of the sample mean? Specify the mean and standard deviation?
Mean of y̅ =266 SD(y̅ )= 16/√(60)=2.066
f) What’s the probability that the mean duration of these patients’ pregnancies will be less than
260 days?
Normalcdf(-9^99,260,266,2.066)=0.0018
6. Carbon monoxide (CO) emissions for a certain kind of car vary with mean 2.9g/mi and
standard deviation 0.4 g/mi. A company has 80 of these cars in its fleet. Let y (bar) represent the
mean CO level for the company’s fleet.
a) What’s the approximate model for the distribution of y (bar)?
y̅ ~ N(2.9,0.4/√ 80) i.e. y̅ ~ N(2.9,0.0447)
b) Estimate the probability that y (bar) is between 3.0 and 3.1 g/mi?
Normalcdf(3,3.1,2.9,0.0447)=0.0126
c) There is only 5% chance that the fleet’s mean CO level is greater than what value?
InvNorm(0.95,2.9,0.0447)=2.97