A - F , please show work . The number of people who shop at a particular grocery
ID: 3364718 • Letter: A
Question
A - F , please show work . The number of people who shop at a particular grocery store from 8 am to 9 am has an unknown probability distribution with a mean of 100 and a variance of 100.
A) what are the mean and standard deviation of the distribution .
B) Whaqt is the likelihood that less than 115 people will show up at the store between 8 and 9 tomorrow?
C) If the distribition were normal , what are the mean and standard deviation of the distribution ?
D)If the distribution were normal , what is the likelhood that less than 115 people will show up at teh store between 8 and 9 tomorrow?
E)Consider 100 days selected at random and again suppose the distribution of the number of people who vistit the store were unknown. What are the mean and standaqrd deviation of the average number of people who visit the store over those hours?
F) What is the likelihood that an average of less than 115 people will show up at the store between 8 and 9 over the 100 day period ?
Explanation / Answer
a) here mean =100 customers/hour
std deviation =(100)1/2 =10
b)
as 115 is 1.5 std deviation from mean value.
therefore from Chebychev's likelihood that less than 115 people will show up at the store between 8 and 9 tomorrow
=P(X<115) =1-1/2k2 = 1-1/(2*1.52) = 0.7778
c)
for normal distribution:
mean =100 customers/hour
std deviation =(100)1/2 =10
d)
likelhood that less than 115 people will show up at teh store between 8 and 9 tomorrow=P(X<115)
=P(Z<(115-100)/10)=P(Z<1.5)=0.9332
e)
from central limit theorum: mean =100
and std deviation of sample mean =std deviation of population/(n)1/2=10/(100)1/2 =1
f) likelihood that an average of less than 115 people will show up at the store between 8 and 9 over the 100 day period =P(X<115) =P(Z<(115-100)/1)=P(Z<15) =~1.0000
(please revert for any clarification required)
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