Which of the following values is closest to the standard deviation for the set

Exhibit B: A restaurant has tracked the number of meals served at lunch over the last four weeks. The data shows little in terms

Mekhanik [1.2K]

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!!!

8 0

3 years ago

A certain company sends 40% of its overnight mail parcels by means of express mail service A1. Of these parcels, 4% arrive after

Harrizon [31]

Answer:

(a) The probability that a randomly selected parcel arrived late is 0.026.

(b) The probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c) The probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d) The probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

Step-by-step explanation:

Consider the tree diagram below.

(a)

The law of total probability sates that: $P(A)=P(A|B)P(B)+P(A|B')P(B')$

Use the law of total probability to determine the probability of a parcel being late.

$P(L)=P(L|A_{1})P(A_{1})+P(L|A_{2})P(A_{2})+P(L|A_{3})P(A_{3})\\=(0.04\times0.40)+(0.01\times0.50)+(0.05\times0.10)\\=0.026$

Thus, the probability that a randomly selected parcel arrived late is 0.026.

(b)

The conditional probability of an event A provided that another event B has already occurred is:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:

$P(A_{1}|L)=\frac{P(L|A_{1})P(A_{1})}{P(L)} \\=\frac{0.04\times 0.40}{0.026} \\=0.615$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{2}|L)=\frac{P(L|A_{2})P(A_{2})}{P(L)} \\=\frac{0.01\times 0.50}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{3}|L)=\frac{P(L|A_{3})P(A_{3})}{P(L)} \\=\frac{0.05\times 0.10}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

4 0

3 years ago

PLS HELP ME ASAP THIS IS DUE IN 20 MINUTES PLSSSSSS

vichka [17]

Answer:

C

Step-by-step explanation:

sailboat A was going 8 ft per second, 40/5=8

sailboat B was going 7 ft per second, 56/8=7

7 0

3 years ago

Read 2 more answers

What is the outlier of the set of data? 33, 46, 41, 41, 30, 9, 14, 33, 15, 26, 12, 46, 28, 36, 95 ____________________

Delvig [45]

The outlier is 95.

All the numbers are from 9 - 46 EXCEPT for 95

Answer = 95

Hope this helped :)

7 0

3 years ago

Which reason could be used for statement 3??

Basile [38]

Option 3, definition of congruent segments is the answer I'm pretty sure

4 0

2 years ago