Can someone help on this? Please :)
2 answers:
answer:
N(t)=2×
17
15
step by step explain:
before lesson (t=0), she knows 2 words.
after a week (t=1), she knows
2×(1+70%)=3.4 words
after one more week (t=2), she knows
3.4×(1+70%)=5.78 words
one more week later (t=3), she knows
5.78×(1+70%)=9.826 words
and so on ...
from pattern shown above, we know that she knows
2× words after t weeks
so N(t)=2×
after 4 weeks (t=4), she knows 2×=16.7042≈17 words
for learning 5000 words, she need:
2×=5000
=2500
t log(1.7)=log(2500)
t=
=14.7448727
≈15 (round up)
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Answer:
a) There is a 69.155 probability that a randomly-selected customer will take less than 3 minutes.
b) There is a 76.11% probability that the average time of two randomly-selected customers will take less than 3 minutes.
c) There is a 90.82% probability that the average time of 64 randomly-selected customers will take less than 2.8 minute.
d) There is a 93.19% probability that the average time of 81 randomly-selected customers will take less than 2.8 minutes.
e) There is a 99.38% probability that the average time of 225 randomly-selected customers will take less than 2.8 minutes.
f) There is a 99.95% probability that the average time of 400 randomly-selected customers will take less than 2.8 minutes.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
The mean checkout time in the express lane of a large grocery store is 2.7 minutes, and the standard deviation is 0.6 minutes. This means that .
(a) What is the probability that a randomly-selected customer will take less than 3 minutes?
This is the pvalue of Z when . So:
has a pvalue of 0.6915.
There is a 69.155 probability that a randomly-selected customer will take less than 3 minutes.
(b) What is the probability that the average time of two randomly-selected customers will take less than 3 minutes?
We are now working with the average of a sample, so we must use instead of
in the formula of Z.
has a pvalue f 0.7611.
There is a 76.11% probability that the average time of two randomly-selected customers will take less than 3 minutes.
(c) The probability that the average time of 64 randomly-selected customers will take less than 2.8 minutes is
This is the pvalue of Z when
has a pvalue of 0.9082.
There is a 90.82% probability that the average time of 64 randomly-selected customers will take less than 2.8 minute.
(d) The probability that the average time of 81 randomly-selected customers will take less than 2.8 minutes is.
has a pvalue of 0.9319.
There is a 93.19% probability that the average time of 81 randomly-selected customers will take less than 2.8 minutes.
(e) The probability that the average time of 225 randomly-selected customers will take less than 2.8 minutes is.
has a pvalue of 0.9938.
There is a 99.38% probability that the average time of 225 randomly-selected customers will take less than 2.8 minutes.
(f) The probability that the average time of 400 randomly-selected customers will take less than 2.8 minutes is.
has a pvalue of 0.9995.
There is a 99.95% probability that the average time of 400 randomly-selected customers will take less than 2.8 minutes.