Download Simple questions part 1

Simple questions part 1
4.
(1) Assume that the readings on the thermometers are normally distributed with a mean
of 0o and standard deviation of 1.00o C. A thermometer is randomly selected and tested.
Draw a diagram and find the probability in degrees.
Between -1.18 and 2.15
Answer:
Mean = 0oCnd SD = 1oC
Probability of temperature between -1.18o and 2.15o
 P( -1.18 < X < 2.15) = P(z1 < Z < z2)
 1.18  0
 1.18
1
2.15  0
z2 
 2.15
1
 P(1.18  Z  2.15)
 0.3810  0.4842  0.8652
z1 
2) The lengths of pregnancies are normally distributed with a mean of 268 days and a
standard deviation of 15 days. A woman claims to give birth 308 after conception.
a. Find probability of a pregnancy lasting 308 days or longer. What does the result
suggest?
b. If we stipulate that a premature birth is in the lowest 4%, find the length that separates
premature from full term births.
Answer:
a)
Mean duration of pregnancy:= 268 days, with SD = 15days
P(X  308) = P(Zz)
308  268
 2.6667
15
 P( Z  2.667)  0.5  0.4962  0.0038
z
 P( X  308days )  0.0038
The result suggests that there is very little possibility that the pregnancy duration will be
more than or equal to 308days.
b)
P(Z = z) = 4% = 0.04
Where z = 0.10
X  268
0.10 
 1.5  X  268
15
X  268  0.10  268.10
Hence the actual number of days = 268.10 and the expected number of days = 268.
Their difference gives the desired value.
(3) Ages 62, 46, 68, 64, 57. Assume the samples of size 2 are randomly selected with
replacement from the population of five ages.
a. after identifying the 25 different possible samples, find the mean of each of them.
b. Describe sampling distribution of means.
c. Find the mean of the sampling distribution
d. Is the mean of the sampling distribution (from c.) equal to the mean of the five ages
above? Are those always equal?
Answer:
a) probability of selecting 2 samples = 2/5 (with replacement)
Hence probability of selecting a sample = 1/5
b) Since replacement is allowed, the repetition of choosing a sample is possible with
samples n = 25, and p = 1/5
Such that mean = np = 25/5 = 5
c) Mean for the given samples = 59.4
d) No the means are not equal. However, if the samples are repeated and replaced
every-time, then there is chance for the two samples to be equal.
(4) Use the Central Limit Theorem - Assume normal distribution with a mean of 172
lb and standard deviation of 29 lb
a. 4 men are randomly selected, what is the probability that they have a mean weight
between 160 and 180 lb.
b. Why can the central limit theorem be used in (a.), even though the sample size does
not exceed 30?
Answer:
Mean = 172lb, SD = 29lb
P( 160lbs  X  180lbs)  P( z1  Z  z 2)
160  172
 0.8275
29 / 4
180  172
 0.5517
a) z 2 
29 / 4
 P(0.8275  Z  0.5517)
z1 
 0.295  0.2088  0.5038
b) Since the distribution of data is assumed to be normally distributed, we can use the
central limit theorem, with the fact that for any X ~ N(mean, variance), the corresponding
Z~N(0,1) with z = (X-mean)/SD.
(5) Building a bench to seat 18 men, normal distribution with a mean of 14.4 in and std
deviation of 1.0 in hip breadth
a. What is the minimum length of the bench if you want a 0.975 probability that it will fit
the combined hip breadths of 18 randomly selected men?
b. What would be wrong with actually using the result form part (a.) as the bench length?
Answer: n = 18men, mean = 14.4inches, with SD = 1.0inch
a) P(X = x) = P(Z =z) = 0.975

xx
 P(2.24 
)  0.975
s/ n
x  14.4
 P(2.24 
)  0.975
1 / 18
Thus we have:
x  14.4
2.24 =
= 2.24 *1.451 = x – 14.4
1 / 18
3.25 = x – 14.4 => x = 3.25 +14.4 = 17.65
Hence the minimum length of the bench should be 17.65inches.
b)
If we wrongly use the value 17.65, the deviation in length will be more and hence the
required probability will be small.
(6) Estimate the probability of getting at least 65 girls in 100 births. Assume boys and
girls are equally likely. Is it unusual to get at least 65 girls in 100 births?
Answer:
p = 65/100 = 0.65, q = 0.35, n = 100
Hence P = 0.50 = Q (Boys and girls are equally likely)
P( X  65 girls )  P( Z  z )
X  np
0.65  0.5
where : z 

 3.144
npq
0.35 * 0.65 / 100
P( Z  3.144)  0.5  P(0  Z  3.144)  0.5  0.4992  0.0008
Hence the percentage of at-least 65% of female birth is very small, hence it is it unusual
to get at least 65 girls in 100 births.
(7) Construct Normal Quantile Plots, identify if the corresponding z scores that are used
for a normal quantile plot the construct the normal Qplot and determine whether the data
appear to be from a population with normal distribution.
a. LA Lakers heights, use the sample in inches 85, 79, 82, 73, 78
Answer: fi = (i -3/8) / (n+1/4)
i
xi
fi
1
2
3
4
5
73
78
79
82
85
0.5
0.722
0.8076
0.8529
0.8809
Quantile plot:
Data values, x
Quantile - Plot
86
84
82
80
Series1
78
76
74
72
0
0.2
0.4
0.6
0.8
1
fraction , f
Normal quantile Plot:
Normal Q-Q Plot
Expected Normal
Value
86
84
82
80
78
76
74
72.5 75.0 77.5 80.0 82.5 85.0
Observed Value
5.
(1) Find critical value zo/2 that corresponds with the given confidence level of 92%
Answer:
For 92% confidence level, the critical z-value = 1.75
i.e. when  =0.08level, z-value =1.75 (Look for the z-value from Normal tables).
(2) Find the margin of error; assume the sample is used to estimate a population
proportion. Find the margin for E.
a. n=1200, x=400, 99% confidence
Answer:
Margin of error = Z-score* standard error
For 99% confidence level, Z-score = 2.575
Proportion p = x/n = 400/1200 = 1/3
q = 1- p = 1-1/3 = 2/3
Hence standard error for p: = SE (p) =
pq
=
n
1/ 3 * 2 / 3
 0.0136
1200
So, the margin of error: 2.575 * 0.0136 = 0.03502
(3) Of 491 randomly selected adults 65% favor the death penalty
a. find the point estimate of the % of adults who are in favor of the death penalty.
Answer:
N = 491, x = 65% of 491 => 319.15,
I.e., p = x/N = 319.15/491 = 0.65, q = 1-p = 1-0.65 = 0.35
Then the mean => np = 491*0.65 = 319.15
Hence the point estimate is 319.65 for 65% of adults who are in favor of the death
penality.
b. find a 95% confidence interval estimate of the % of adults who are in favor.
Answer:
95% confidence interval:
pq
0.65 * 0.35
 0.65  1.96 *
n
491
 0.65  1.96 * 0.0215
 0.65  0.04214
 (0.60786,69214)
This is the required confidence interval estimate for 65% of adults who are in favor.
p  Z *
c. can we safely conclude that the majority of adults are in favor? Explain why.
Answer: Since the percentage of adults who are in favor of death penalty is 65% than that
of young ones, we conclude that the majority of adults are in favor for the capital
punishment.(death penalty)
(4) Women heights are normally distributed with a mean of 63.6 in. and a std deviation of
2.5 in.
a. how many women must be surveyed if we want to estimate the % who are taller than
5ft? assume that we want 98% confidence that the error is no more that 2.5 % points.
(answer is substantially smaller than 2712)
Answer:
We have to compute the sample size ‘n’.
To find this n, we have the formula given as:
Z2 *S2
n  2
E
Where
Z   2.33
S  2.5  S 2  6.25
And
E  2.5%
Then,
Z 2 * S 2 (2.33) 2 * (6.25) 5.4289 * 6.25


 2712.45 ~ 2712
E2
(2.5%) 2
(0.025) 2
Hence the number of women to be surveyed should be not more than 2712.
n
(5) Margin of error and confidence Int.,
a. ages of drivers in the passing lane while driving 25 mi/h with left signal flashing: 99%
confidence; n=50, mean=80.5 yrs and std is 4.6 yrs
Answer:
M arg in  of  error :
s
4.6
Z *
 2.575 *
 1.675
n
50
Confidence  int erval :
mean  M arg in  of  error
80.5  1.675
(78.825,82.175)
This is the required 99% confidence interval for the true mean age for drivers.
(6) Find confidence Int.,
a. 99% confidence; n=15, mean=496, s=0.07in
Answer:
M arg in  of  error :
s
0.07
t ( n1, / 2) *
 2.977 *
 0.0538
n
15
Confidence  int erval :
mean  M arg in  of  error
496  0.0538
(495.9462,496.0528)
This is the required 99% confidence interval for the true mean.