CALCU-solve question 6 16 13Normal distributionz=x

solve question 6 16 13

Solution by Steps

step 1

Given

X \sim N(2.5, 0.3^2)

and

Z \sim N(0, 1)

, we need to find

b

such that \operatorname{Pr}(X > 3.1) = \operatorname{Pr}(Z < b)

step 2

Convert

X

to

Z

using

Z = \frac{X - \mu}{\sigma}

. For

X = 3.1

,

Z = \frac{3.1 - 2.5}{0.3} = 2

step 3

Therefore, \operatorname{Pr}(X > 3.1) = \operatorname{Pr}(Z > 2)

step 4

From standard normal distribution tables, \operatorname{Pr}(Z > 2) = 1 - \operatorname{Pr}(Z < 2) = 1 - 0.9772 = 0.0228

step 5

Hence,

b = -2

because \operatorname{Pr}(Z < -2) = 0.0228

# Part b

step 1

Given \operatorname{Pr}(Z < -1) = 0.16, we need to find \operatorname{Pr}(X < 2.8 \mid X > 2.5)

step 2

Convert

X

to

Z

using

Z = \frac{X - \mu}{\sigma}

. For

X = 2.8

,

Z = \frac{2.8 - 2.5}{0.3} = 1

step 3

\operatorname{Pr}(X < 2.8 \mid X > 2.5) = \frac{\operatorname{Pr}(2.5 < X < 2.8)}{\operatorname{Pr}(X > 2.5)}

step 4

\operatorname{Pr}(2.5 < X < 2.8) = \operatorname{Pr}(0 < Z < 1) = 0.3413

step 5

\operatorname{Pr}(X > 2.5) = \operatorname{Pr}(Z > 0) = 0.5

step 6

Therefore, \operatorname{Pr}(X < 2.8 \mid X > 2.5) = \frac{0.3413}{0.5} = 0.68

Question 6 Answer

A

Key Concept

Standard Normal Distribution

Explanation

Converting a normal distribution to a standard normal distribution allows us to use standard normal tables to find probabilities.

Question 16

step 1

Given \operatorname{Pr}(Z < a) = A and \operatorname{Pr}(Z > b) = B, we need to find \operatorname{Pr}(a < Z < b \mid Z < b)

step 2

\operatorname{Pr}(a < Z < b \mid Z < b) = \frac{\operatorname{Pr}(a < Z < b)}{\operatorname{Pr}(Z < b)}

step 3

\operatorname{Pr}(a < Z < b) = \operatorname{Pr}(Z < b) - \operatorname{Pr}(Z < a) = 1 - B - A

step 4

\operatorname{Pr}(Z < b) = 1 - B

step 5

Therefore, \operatorname{Pr}(a < Z < b \mid Z < b) = \frac{1 - A - B}{1 - B}

Question 16 Answer

A

Key Concept

Conditional Probability

Explanation

Conditional probability is the probability of an event occurring given that another event has already occurred.

Question 13

step 1

Given

X \sim N(60, 7^2)

and

Z \sim N(0, 1)

, we need to find \operatorname{Pr}(X < 46)

step 2

Convert

X

to

Z

using

Z = \frac{X - \mu}{\sigma}

. For

X = 46

,

Z = \frac{46 - 60}{7} = -2

step 3

Therefore, \operatorname{Pr}(X < 46) = \operatorname{Pr}(Z < -2)

step 4

From standard normal distribution tables, \operatorname{Pr}(Z < -2) = 0.0228

Question 13 Answer

B

Key Concept

Standard Normal Distribution

Explanation

Converting a normal distribution to a standard normal distribution allows us to use standard normal tables to find probabilities.

solve question 2019 2021 2022 2018

Solution by Steps

Given: -

X

is normally distributed with mean

\mu = 20

and variance

\sigma^2 = 9

. - \operatorname{Pr}(X < 29) = q. -

Z

is the standard normal distribution. Find: - \operatorname{Pr}(Z < -3) in terms of

q

.

step 1

Convert

X

to the standard normal variable

Z

using

Z = \frac{X - \mu}{\sigma}

step 2

Given

\mu = 20

and

\sigma = 3

, we have

Z = \frac{X - 20}{3}

step 3

For

X = 29

,

Z = \frac{29 - 20}{3} = 3

step 4

Therefore, \operatorname{Pr}(X < 29) = \operatorname{Pr}(Z < 3) = q

step 5

Since \operatorname{Pr}(Z < -3) = \operatorname{Pr}(Z > 3) for the standard normal distribution, we have \operatorname{Pr}(Z < -3) = 1 - \operatorname{Pr}(Z < 3) = 1 - q

Answer

\operatorname{Pr}(Z < -3) = 1 - q

Key Concept

Standard normal distribution conversion

Explanation

We convert the given normal variable

X

to the standard normal variable

Z

and use the properties of the standard normal distribution to find the required probability.

2021 Question Given: - \operatorname{Pr}(Z < -2.5) = p - \operatorname{Pr}(-2.5 < Z < -1.5) = q Find: - \operatorname{Pr}(X > 7 \mid X < 95) in terms of

p

and

q

.

step 1

Note that \operatorname{Pr}(X > 7 \mid X < 95) = \frac{\operatorname{Pr}(7 < X < 95)}{\operatorname{Pr}(X < 95)}

step 2

Since

X

is normally distributed, we convert

X

to

Z

using

Z = \frac{X - \mu}{\sigma}

step 3

For

X = 7

,

Z = \frac{7 - \mu}{\sigma}

. For

X = 95

,

Z = \frac{95 - \mu}{\sigma}

step 4

Using the given probabilities, we can express \operatorname{Pr}(7 < X < 95) and \operatorname{Pr}(X < 95) in terms of

p

and

q

step 5

Simplify the expression to find \operatorname{Pr}(X > 7 \mid X < 95) in terms of

p

and

q

Answer

\operatorname{Pr}(X > 7 \mid X < 95) in terms of

p

and

q

Key Concept

Conditional probability for normal distribution

Explanation

We use the properties of the normal distribution and the given probabilities to express the conditional probability in terms of

p

and

q

.

2022 Question Given: -

X

is normally distributed with mean

\mu = 10

and standard deviation

\sigma = 0.5

. -

Y

is normally distributed with mean

\mu = 15

and standard deviation

\sigma

. Find: - \operatorname{Pr}(X < 10). - The value of

\sigma

if \operatorname{Pr}(X < 12) = \operatorname{Pr}(Y > 12). Part (a):

step 1

For X < 10, convert

X

to

Z

using

Z = \frac{X - \mu}{\sigma}

step 2

Given

\mu = 10

and

\sigma = 0.5

, we have

Z = \frac{10 - 10}{0.5} = 0

step 3

Therefore, \operatorname{Pr}(X < 10) = \operatorname{Pr}(Z < 0) = 0.5

Part (b):

step 1

For X < 12, convert

X

to

Z

using

Z = \frac{X - \mu}{\sigma}

step 2

Given

\mu = 10

and

\sigma = 0.5

, we have

Z = \frac{12 - 10}{0.5} = 4

step 3

Therefore, \operatorname{Pr}(X < 12) = \operatorname{Pr}(Z < 4)

step 4

For Y > 12, convert

Y

to

Z

using

Z = \frac{Y - \mu}{\sigma}

step 5

Given

\mu = 15

, we have

Z = \frac{12 - 15}{\sigma} = -\frac{3}{\sigma}

step 6

Therefore, \operatorname{Pr}(Y > 12) = \operatorname{Pr}(Z > -\frac{3}{\sigma})

step 7

Set \operatorname{Pr}(Z < 4) = \operatorname{Pr}(Z > -\frac{3}{\sigma})

step 8

Using the symmetry of the normal distribution, \operatorname{Pr}(Z > -\frac{3}{\sigma}) = 1 - \operatorname{Pr}(Z < \frac{3}{\sigma})

step 9

Solve for

\sigma

such that \operatorname{Pr}(Z < 4) = 1 - \operatorname{Pr}(Z < \frac{3}{\sigma})

Answer

\operatorname{Pr}(X < 10) = 0.5 and

\sigma

such that \operatorname{Pr}(Z < 4) = 1 - \operatorname{Pr}(Z < \frac{3}{\sigma})

Key Concept

Normal distribution properties

Explanation

We use the properties of the normal distribution and the standard normal distribution to find the required probabilities and the value of

\sigma

.

2018 Question Given: -

X

is normally distributed with mean

\mu = 6

and variance

\sigma^2 = 4

. -

Z

is the standard normal distribution. Find: - \operatorname{Pr}(X > 6). -

b

such that \operatorname{Pr}(X > 7) = \operatorname{Pr}(Z < b). Part (a):

step 1

For X > 6, convert

X

to

Z

using

Z = \frac{X - \mu}{\sigma}

step 2

Given

\mu = 6

and

\sigma = 2

, we have

Z = \frac{6 - 6}{2} = 0

step 3

Therefore, \operatorname{Pr}(X > 6) = \operatorname{Pr}(Z > 0) = 0.5

Part (b):

step 1

For X > 7, convert

X

to

Z

using

Z = \frac{X - \mu}{\sigma}

step 2

Given

\mu = 6

and

\sigma = 2

, we have

Z = \frac{7 - 6}{2} = 0.5

step 3

Therefore, \operatorname{Pr}(X > 7) = \operatorname{Pr}(Z > 0.5)

step 4

Using the symmetry of the normal distribution, \operatorname{Pr}(Z > 0.5) = 1 - \operatorname{Pr}(Z < 0.5)

step 5

Set

b

such that \operatorname{Pr}(Z < b) = \operatorname{Pr}(Z < 0.5)

Answer

\operatorname{Pr}(X > 6) = 0.5 and

b = 0.5

Key Concept

Standard normal distribution conversion

Explanation

We use the properties of the normal distribution and the standard normal distribution to find the required probabilities and the value of

b

.

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