<span>Defective rate can be expected
to keep an eye on a Poisson distribution. Mean is equal to 800(0.02) = 16,
Variance is 16, and so standard deviation is 4.
X = 800(0.04) = 32, Using normal approximation of the Poisson distribution Z1 =
(32-16)/4 = 4.
P(greater than 4%) = P(Z>4) = 1 – 0.999968 = 0.000032, which implies that
having such a defective rate is extremely unlikely.</span>
<span>If the defective rate in the
random sample is 4 percent then it is very likely that the assembly line
produces more than 2% defective rate now.</span>
A. 7.5 pages per hour B. $2.66 for one page
Answer:
Option E is correct.
The expected number of meals expected to be served on Wednesday in week 5 = 74.2
Step-by-step Explanation:
We will use the data from the four weeks to obtain the fraction of total days that number of meals served at lunch on a Wednesday take and then subsequently check the expected number of meals served at lunch the next Wednesday.
Week
Day 1 2 3 4 | Total
Sunday 40 35 39 43 | 157
Monday 54 55 51 59 | 219
Tuesday 61 60 65 64 | 250
Wednesday 72 77 78 69 | 296
Thursday 89 80 81 79 | 329
Friday 91 90 99 95 | 375
Saturday 80 82 81 83 | 326
Total number of meals served at lunch over the 4 weeks = (157+219+250+296+329+375+326) = 1952
Total number of meals served at lunch on Wednesdays over the 4 weeks = 296
Fraction of total number of meals served at lunch over four weeks that were served on Wednesdays = (296/1952) = 0.1516393443
Total number of meals expected to be served in week 5 = 490
Number of meals expected to be served on Wednesday in week 5 = 0.1516393443 × 490 = 74.3
Checking the options,
74.3 ≈ 74.2
Hence, the expected number of meals expected to be served on Wednesday in week 5 = 74.2
Hope this Helps!!!
Well that symbol is an indention or multiply sign of expression
0.50
imagine it like a dollar.
1/4 of a dollar is 0.25
meaning 2/4 is 0.50
