Suppose a research firm conducted a survey to determine the average amount of money steady smokers spend on cigarettes during a week. a sample of 100 steady smokers revealed that the sample mean is $20 and the sample standard deviation is $5. what is the probability that a sample of 100 steady smokers spend between $19 and $21?
a.0.4772
b.0.0228
c.0.9544
d.$20

option (b) 0.0228

Step-by-step explanation:

Data provided in the question:

Sample size, n = 100

Sample mean, μ = $20

Standard deviation, s = $5

Confidence interval = between $19 and $21

Now,

Confidence interval = μ ±

thus,

Upper limit of the Confidence interval = μ +

or

$21 =  $20 +

or

z = 2

Now,

P(z = 2) = 0.02275                       [From standard z vs p value table]

or

P(z = 2) ≈ 0.0228

Hence,

the correct answer is option (b) 0.0228

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?