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sweet-ann [11.9K]

3 years ago

13

The relative frequency distribution of the number of phobias reported by a hypothetical sample of 500 college students is given

as follows.
0–2 0.48
3–5 0.26
6–8 0.12
9–11 0.09
12–14 0.05

Required:
a. What is the probability that a college student expresses fewer than three phobias?
b. What is the probability that a college student expresses more than eight phobias?
c. What is the probability that a college student has between 3 and 11 phobias?

1 answer:

goldenfox [79]3 years ago

3 0

Answer:

a. 0.48

b. 0.14

c. 0.47

Step-by-step explanation:

Data provided in the question

0 - 2 0.48

3 - 5 0.26

6 - 8 0.12

9- 11 0.09

12- 14 0.05

Based on the above information

a. The probability for fewer than three phobias is

= P( x < 3)

= 0.48

b. The probability for more than eight phobias is

= P( x >8)

= 0.09 + 0.05

= 0.14

c. Probability between 3 and 11 phobias is

= P(3 < x < 11)

= 0.26 + 0.12 + 0.09

= 0.47

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Answer:

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

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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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3 years ago