Question

A normal population has mean μ = 31 and standard deviation σ = 6. (a) What proportion of the population is between 19 and 28? (b) What is the probability that a randomly chosen value will be between 26 and 36? Round the answers to at least four decimal places. Part 1 of 2 The proportion of the population between 19 and 28 is . Part 2 of 2 the probability that a randomly chosen value will be between 26 and 36 is?

Answer #1

Solution :

(a)

P(19 < x < 28) = P[(19 - 31)/ 6) < (x - ) / < (28 - 31 ) / 6) ]

= P(1.33 < z < -0.5)

= P(z < -0.5) - P(z < 1.33)

= 0.5997

**Proportion = 0.5997**

(b)

(26 < x < 36) = P[(26 - 31)/ 6) < (x - ) / < (36 - 31 ) / 6) ]

= P(-0.83 < z < 0.83)

= P(z < 0.83) - P(z < -0.83)

= 0.5935

**Proportion = 0.5935**

A normal population has mean μ 30 = 31 and standard deviation σ=
7
(a) What proportion of the population is between 15 and 25?
(b) What is the probability that a randomly chosen value will be
between 25 and 35?
Round the answers to at least four decimal places.

A normal population has mean μ= 34 and standard deviation σ=
10.
(a) What proportion of the population is between 10 and 20?
(b) What is the probability that a randomly chosen value will be
between 28 and 38?
Round the answers to at least four decimal places

A normal population has mean μ =40 and standard deviation σ
=9.
(a) What proportion of the population is between 20 and 30?
(b) What is the probability that a randomly chosen value will be
between 35 and 45?
Round the answers to at least four decimal places.

A normal population has mean μ=40 and standard deviation σ
=9.
(a) What proportion of the population is between 20 and 30 ?
(b) What is the probability that a randomly chosen value will be
between 35 and 45?
Round the answers to at least four decimal places

A normal population has mean μ = 9 and standard
deviation σ = 6.
a.
What proportion of the population is less than 20?
b.
What is the probability that a randomly chosen value will be
greater than 5?

normal population has mean =μ10 and standard deviation =σ7. (a)
What proportion of the population is less than 18? (b) What is the
probability that a randomly chosen value will be greater than 3?
Round the answers to four decimal places. Part 1 of 2 The
proportion of the population less than 18 is . Part 2 of 2 The
probability that a randomly chosen value will be greater than 3 is
.

A normal population has mean =μ9 and standard deviation =σ5.
(a) What proportion of the population is less than 20?
(b) What is the probability that a randomly chosen value will be
greater than 5?
Round the answers to four decimal places.

A normal population has mean
μ
=
52
and standard deviation
σ
=
7
. What is the
82
nd percentile of the population?

A population is normally distributed with mean μ = 100 and
standard deviation σ = 20. Find the probability that a value
randomly selected from this population will have a value between 90
and 130. (i.e., calculate P(90<X<130))

A population of values has a normal distribution with μ = 53.9
and σ = 17.9 . You intend to draw a random sample of size n = 28 .
Find the probability that a single randomly selected value is
between 57.6 and 58.3. P(57.6 < X < 58.3) = Find the
probability that a sample of size n = 28 is randomly selected with
a mean between 57.6 and 58.3. P(57.6 < M < 58.3) = Enter your
answers...

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