Download Normal distribution – exercise: Answers

Normal distribution – exercise:
Answers
1.
i.
0.227 (or 22.7%; allow answers in decimal or percentage form throughout)
ii.
13
iii.
30.3 s – 61.7 s
iv.
Mean = 34.1s, standard deviation = 14.5 s
v.
The claim is justified; the mean is lower, but the standard deviation is much larger.
i.
0.706
ii.
About 4 days, depending on the number of shopping days chosen for the month.
Allow 14.7% of ‘their number of days’, provided the choice is reasonable.
iii.
Any reasonable suggestions; for example, increased sales during the Christmas
shopping period, decreased sales during summer.
iv.
Days where sales are £4,392 or lower.
2.
3.
i.
0.1
ii.
13.1%
iii.
0.148 s – 0.182 s
iv.
0.11%
4.
i.
0.1
ii.
25 fish.
iii.
Mean of sample = 22.3 cm, standard deviation = 3.54 cm.
The fish in Lake B are shorter than those in Lake A on average, and their lengths
are less varied.
5.
i.
Any three valid features. For example: bell-shaped curve, symmetrical about the
mean, asymptotic to z-axis as z → ± ∞, about two-thirds of population within 1
standard deviation of the mean, almost all of the population within 3 standard
deviations of the mean.
ii.
If the dance students were physically fitter than the broader student population,
their recovery times might be shorter. This would tend to skew the overall results
(i.e. a negative skew).
iii.
Stem-and-leaf diagram with following parameters: Min = 67, LQ = 78, Median =
87, UQ = 92, Max = 116 (no outliers).
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iv.
Possibly: quartiles are not symmetrically distributed around the median, but
maximum and minimum values are more symmetrical, with no obvious outliers.
v.
Mean = 87.1 s, standard deviation = 12.5 s.
vi.
Number of people within 1 s.d. of the mean (i.e. in the range 74.6 s – 100.4 s) is 24
out of 35, or about 68.6%. For a normal distribution, the percentage of the
population in the range z = ±1 is 68.3%. This supports the idea that a normal
distribution is a reasonable model.
6.
i.
The values given correspond to z-values of ±1; for a normal distribution,
approximately two-thirds of the population is within one standard deviation of the
mean.
ii.
Mean ± 1.96 s.d. = 49.5 cm ± 1.96 × 1.05 cm = 47.3 cm – 51.6 cm.
iii.
The values given correspond to z-scores of -0.33 and 3.67. The percentage of the
normal distribution in this range would be (50 + 12.9 – 0.11)% = 62.8%.
iv.
We need a z-score of -1.28, which corresponds to a head circumference of
(49.5 – 1.28 × 1.5) cm = 47.1 cm.
i.
Because half the filled bottles would contain less than the stated contents.
ii.
z-score = (1020 – 1015) ÷ 4.6 = 1.09. Probability that contents exceed 1020 ml  =
13.8%.
iii.
‘Under-filled’ means contents less than 1000 ml. This corresponds to a z-score
of -15 ÷ 4.6 = 3.26. The probability of a bottle being under-filled is less than 1 in
1000; many normal distribution tables only go to z = 3.09, which gives a probability
p(Z > 3.09) of 0.1%.
iv.
Mean + 2.33 s.d. = (1015 + 2.33 × 4.6) ml = 1025.7 ml
v.
Stem-and-leaf diagram with following parameters: Min = 1004, LQ = 1018.75,
Median = 1021.5, UQ = 1027.25, Max = 1036 (no outliers).
vi.
Mean = 1021.7 ml, standard deviation = 7.7 ml.
vii.
The mean and standard deviation have both increased, which will lead to more of
the extreme high values.
7.
8.
i.
A delay of 15 minutes corresponds to a z-score of 3 ÷ 2.5 = 1.2.
The probability p(Z > 1.2) = 0.115.
ii.
We might expect both the mean and standard deviation of delays at an
international airport to be greater than those at a domestic airport. International
flights may be subject to a wider variety of factors causing delays, such as late
arrivals of incoming flights. With incoming flights coming from around the world,
the chances of occasional extreme delays would be larger.
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