STATISTICS MIXED PROBLEMS
1. Consider a population with µ = 59.6 and σ = 3.45.
̅ = 60.7 from a sample of size 15.
(A) Calculate the z-score for 𝑿
𝒛 − 𝒔𝒄𝒐𝒓𝒆 =
̅−µ
𝑿
𝝈/√𝒏
=
𝟔𝟎. 𝟕 − 𝟓𝟗. 𝟔
𝟑. 𝟒𝟓/√𝟏𝟓
= 𝟏. 𝟐𝟑𝟓
(B) Could this z-score be used in calculating probabilities
using Table 3 in Appendix B of the text?
Why or why not?
Yes, it can be used to compute probabilities involving the
̅ For instance,
distribution 𝑿.
̅ −µ
𝑿
𝟔𝟎. 𝟕 − 𝟓𝟗. 𝟔
̅
𝑷(𝑿 > 𝟔𝟎. 𝟕) = 𝑷 (
>
) = 𝑷(𝒁 > 𝟏. 𝟐𝟑𝟓) = 𝟎. 𝟏𝟎𝟖𝟒
𝝈/√𝒏
𝟑. 𝟒𝟓/√𝟏𝟓
2. Given a level of confidence of 95% and a population standard
deviation of 9, answer the following:
(A) What other information is necessary to find the sample size(n)?
̅ ± 𝑧0.025 ∙
We know that the 95% confidence interval formula is given by 𝒙
𝜎
, where the part
𝑛
√
𝑬 = 𝒛𝟎.𝟎𝟐𝟓 ∙
𝝈
√𝒏
is the Maximum Error of Estimate. We need to be given this quantity
in order to find out the sample size. Notice that we already know
the values of 𝑧0.025 = 1.96 and 𝜎 = 9.
(B) Find the Maximum Error of Estimate (E) if n = 72.
Show all work.
𝑬 = 𝒛𝟎.𝟎𝟐𝟓 ∙
𝝈
√𝒏
= 𝟏. 𝟗𝟔 ∙
𝟗
√𝟕𝟐
= 𝟐. 𝟎𝟕𝟗
3. A sample of 134 golfers showed that their average score on a
particular golf course was 93.23 with a standard deviation of
6.25.
Answer each of the following (show all work and state the final
answer to at least two decimal places.):
(A) Find the 98% confidence interval of the mean score for all 134
golfers.
̅±
We know that the 98% confidence interval formula is given by 𝒙
𝜎
̅ = 93.23, σ = 6.25 and 𝑧0.01 = 2.326
𝑧0.01 ∙ 𝑛, where we are given that 𝒙
√
So the desired confidence interval is
̅ ± 𝑧0.01 ∙
𝒙
𝜎
√𝑛
= 93.23 ± 2.326 ∙
6.25
√134
= 93.23 ± 1.256
The 98% confidence level interval for the mean score for all 134
golfers is (91.974, 94.486)
(B) Find the 98% confidence interval of the mean score for all
golfers if this is a sample of 110 golfers instead of a sample of
134.
Same computations as above with 134 replaced by 110.
̅ ± 𝑧0.01 ∙
𝒙
𝜎
√𝑛
= 93.23 ± 2.326 ∙
6.25
√110
= 93.23 ± 1.386
The 98% confidence level interval for the mean score for all 110
golfers is (91.844, 94.616)
(C) Which confidence interval is larger and why?
The confidence interval is larger if the number of golfers is
smaller. The reason is that the maximum error of estimate is
larger when the sample is smaller. This makes sense at an
intuitive level as well: the larger the sample, the better it
approximates the population and therefore the smaller the
variance. As a consequence we can be more precise when we work
with a larger sample. That’s why the narrower interval in the case
of more golfers.
4. Assume that the population of heights of female college students
is approximately normally distributed with mean  of 67.26 inches
and standard deviation  of 5.96 inches. A random sample of 78
heights is obtained. Show all work.
(A)
̅ distribution
Find the mean and standard error of the 𝑿
̅ ) = 𝝁 = 𝟔𝟕. 𝟐𝟔
Mean(𝑿
̅ )= 𝝈 = 𝟓.𝟗𝟔 = 𝟎. 𝟔𝟕𝟒𝟖
𝜎(𝑿
√𝒏
(B)
√𝟕𝟖
̅ > 67.50)
Find P(𝑿
̅
̅ > 𝟔𝟕. 𝟓𝟎) = 𝑷 ( 𝑿−𝝁 > 𝟔𝟕.𝟓𝟎−𝟔𝟕.𝟐𝟔) = 𝑷(𝒁 > 𝟎. 𝟑𝟓𝟓𝟔) = 𝟎. 𝟑𝟓𝟗𝟒
𝑷(𝑿
̅)
𝜎(𝑿
𝟎.𝟔𝟕𝟒𝟖
5. The diameters of grapefruits in a certain orchard are normally
distributed with a mean of 5.72 inches and a standard deviation of
0.57 inches. Show all work.
(A) What percentage of the grapefruits in this orchard is larger
than 5.66 inches?
𝑷(𝑫 > 𝟓. 𝟔𝟔) = 𝑷 (
𝑫 − 𝝁 𝟓. 𝟔𝟔 − 𝟓. 𝟕𝟐
>
) = 𝑷(𝒁 > −𝟎. 𝟏𝟎𝟓𝟑) = 𝟎. 𝟓𝟒𝟐
𝝈
𝟎. 𝟓𝟕
About 54.2% of the grapefruits have diameter greater than 5.66
inches
(B) A random sample of 100 grapefruits is gathered and the mean
diameter is calculated. What is the probability that the sample
mean is greater than 5.66 inches?
̅ )= 𝝈 =
𝜎(𝑿
√𝒏
𝟎.𝟓𝟕
√𝟏𝟎𝟎
= 𝟎. 𝟎𝟓𝟕
̅
̅ > 𝟓. 𝟔𝟔) = 𝑷( 𝑿−𝝁 >𝟓.𝟔𝟔−𝟓.𝟕𝟐) = 𝑷(𝒁 > −𝟏. 𝟎𝟓𝟑) = 𝟎. 𝟖𝟓𝟒
𝑷(𝑿
̅)
𝝈(𝑿
𝟎.𝟎𝟓𝟕
With probability of about 85.4% the sample mean
will exceed 5.66 inches
6. A researcher is interested in estimating the noise levels in
decibels at area urban hospitals. She wants to be 98% confident
that her estimate is correct. If the standard deviation is
4.74, how large a sample is needed to get the desired
information to be accurate within 0.69 decibels? Show all work.
At 98% confidence we have that 𝜶 = 𝟎. 𝟎𝟐 and therefore 𝒛𝜶⁄𝟐 = 𝟐. 𝟑𝟐𝟔.
We put the condition
𝐳𝛂⁄𝟐 ∙
𝛔
√𝐧
≤ 𝟎. 𝟔𝟗 → 𝐧 ≥ (𝐳𝛂⁄𝟐 ∙
𝛔
𝟐
) = (𝟐. 𝟑𝟐𝟔 ∙
𝟎.𝟔𝟗
𝟒.𝟕𝟒 𝟐
) = 𝟐𝟓𝟓. 𝟑
𝟎.𝟔𝟗
In conclusion, a sample of size at least 256 is needed to
achieve an accurate estimate within 0.69 decibels with a 98%
confidence level.