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AveGali [126]
1 year ago
13

he marketing department at an online site wants to identify the average age of its users. a sample of 56 uses is selected. the a

verage age in the sample was 22.5 years with a standard deviation of 5 years. assume the population of consumer ages is normally distributed. construct a 90% confidence interval for the average age of all the users of this site.
Mathematics
1 answer:
dolphi86 [110]1 year ago
4 0

The range for the average age of all site visitors is found to be between 21.15 and 23.55.

<h3>Define the term confidence interval?</h3>
  • The likelihood that a quantity will fall between two values near the mean is shown by a confidence interval.
  • Confidence intervals quantify how certain or uncertain a sampling technique is.
  • They are also utilized in regression analysis and hypothesis testing.

For the given question;

Confidence interval  CI = 90% = 0.9

Sample size n = 56 users

Mean (x) = 22.5 years years

Standard deviation (s) = 5 years

Then formula for the CI is given as,

CI = x + z.s/√n

The steps included are-

1. Subtract 1 from sample size to get the degree of freedom df:

df = 56 - 1 = 55

2. To get teh alpha level.

α = (1 - 0.9)/2 = 0.05

3: check df and α in t-distribution to get z score.

z = 2.003

4: To get sample size,

S = s/√n

S = 5/√56

S = 0.66

z x S = 0.66 x 2.033 = 1.35

5: To get the lower range subtract 1.35 from sample mean

22.5  -  1.35 = 21.15

6:n  To get the upper range, add t 1.35 from sample mean.

22.5 + 1.35 = 23.55

Thus, the confidence interval is 21.15 and 23.55.

To know more about the confidence interval, here

brainly.com/question/14825274

#SPJ4

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Step-by-step explanation:

Step 1: Write equation

180 + 8x = 160 + 6x

Step 2: Solve for <em>x</em>

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USA Today reports that about 25% of all prison parolees become repeat offenders. Alice is a social worker whose job is to counse
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Answer:

a) P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039  

P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469  

P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211  

P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422  

P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316

b) E(X) = np = 4*0.75=3

c) Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866

d) P(X \geq 3) \geq 0.98

And the dsitribution that satisfy this is X\sim Binom(n=9,p=0.75

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=4, p=1-0.25=0.75)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039  

P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469  

P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211  

P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422  

P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316

Part b

The expected value is givn by:

E(X) = np = 4*0.75=3

Part c

For the standard deviation we have this:

Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866

Part d

For this case the sample size needs to be higher or equal to 9. Since we need a value such that:

P(X \geq 3) \geq 0.98

And the dsitribution that satisfy this is X\sim Binom(n=9,p=0.75

We can verify this using the following code:

"=1-BINOM.DIST(3,9,0.75,TRUE)" and we got 0.99 and the condition is satisfied.

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