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Generated Graph

Solution by Steps

Given: -

X

is normally distributed with mean

80

and variance

36

. -

Z

is the standard normal random variable. - \operatorname{Pr}(Z<-2.5)=p - \operatorname{Pr}(-2.5<Z<-1.5)=q To find: - \operatorname{Pr}(X>7 \mid X<95) in terms of

p

and

q

.

step 1

Convert

X

to the standard normal variable

Z

using

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

. Here,

\mu = 80

and

\sigma = \sqrt{36} = 6

step 2

For X > 7:

Z = \frac{7 - 80}{6} = \frac{-73}{6}

step 3

For X < 95:

Z = \frac{95 - 80}{6} = \frac{15}{6} = 2.5

step 4

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

step 5

\operatorname{Pr}(7 < X < 95) = \operatorname{Pr}(-\frac{73}{6} < Z < 2.5) = \operatorname{Pr}(Z < 2.5) - \operatorname{Pr}(Z < -\frac{73}{6})

step 6

\operatorname{Pr}(X < 95) = \operatorname{Pr}(Z < 2.5)

step 7

\operatorname{Pr}(Z < 2.5) = 1 - \operatorname{Pr}(Z > 2.5) = 1 - \frac{1}{2} \operatorname{erfc}(\frac{5}{2 \sqrt{2}})

step 8

\operatorname{Pr}(Z < -\frac{73}{6}) \approx 0 (since

\frac{73}{6}

is very large)

step 9

\operatorname{Pr}(7 < X < 95) \approx \operatorname{Pr}(Z < 2.5)

step 10

\operatorname{Pr}(X > 7 \mid X < 95) \approx \frac{\operatorname{Pr}(Z < 2.5)}{\operatorname{Pr}(Z < 2.5)} = 1

Answer

\operatorname{Pr}(X > 7 \mid X < 95) \approx 1

Key Concept

Conditional Probability in Normal Distribution

Explanation

The probability \operatorname{Pr}(X > 7 \mid X < 95) is approximately

1

because the probability of

X

being less than

7

is extremely small compared to the probability of

X

being less than

95

.

For the question from 2022 - insight Q6: Given: -

X

is normally distributed with mean

10

and standard deviation

0.5

. -

Y

is normally distributed with mean

15

and standard deviation

\sigma

. - \operatorname{Pr}(X < 12) = \operatorname{Pr}(Y > 12). To find: - \operatorname{Pr}(X < 10). - The value of

\sigma

. Part (a):

step 1

Convert

X

to the standard normal variable

Z

using

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

. Here,

\mu = 10

and

\sigma = 0.5

step 2

For X < 10:

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

step 3

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

Part (b):

step 1

Convert

X

to the standard normal variable

Z_X

using

Z_X = \frac{X - 10}{0.5}

. For X < 12:

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

step 2

Convert

Y

to the standard normal variable

Z_Y

using

Z_Y = \frac{Y - 15}{\sigma}

. For Y > 12:

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

step 3

\operatorname{Pr}(X < 12) = \operatorname{Pr}(Z_X < 4) \approx 1

step 4

\operatorname{Pr}(Y > 12) = \operatorname{Pr}(Z_Y > -\frac{3}{\sigma}) = 1 - \operatorname{Pr}(Z_Y < -\frac{3}{\sigma})

step 5

Since \operatorname{Pr}(X < 12) = \operatorname{Pr}(Y > 12), we have 1 \approx 1 - \operatorname{Pr}(Z_Y < -\frac{3}{\sigma})

step 6

\operatorname{Pr}(Z_Y < -\frac{3}{\sigma}) \approx 0

step 7

For \operatorname{Pr}(Z_Y < -\frac{3}{\sigma}) \approx 0,

\sigma

must be very large

Answer

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

\sigma

must be very large

Key Concept

Standard Normal Distribution and Conditional Probability

Explanation

The probability \operatorname{Pr}(X < 10) is

0.5

because

10

is the mean of

X

. For \operatorname{Pr}(X < 12) = \operatorname{Pr}(Y > 12),

\sigma

must be very large to make \operatorname{Pr}(Z_Y < -\frac{3}{\sigma}) \approx 0.

For the question from 2018 - Q4: Given: -

X

is normally distributed with mean

6

and variance

4

. -

Z

is the standard normal random variable. To find: - \operatorname{Pr}(X > 6). -

b

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

step 1

Convert

X

to the standard normal variable

Z

using

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

. Here,

\mu = 6

and

\sigma = \sqrt{4} = 2

step 2

For X > 6:

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

step 3

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

Part (b):

step 1

For X > 7:

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

step 2

\operatorname{Pr}(X > 7) = \operatorname{Pr}(Z > 0.5) = 1 - \operatorname{Pr}(Z < 0.5)

step 3

\operatorname{Pr}(Z < 0.5) \approx 0.6915

step 4

\operatorname{Pr}(X > 7) = 1 - 0.6915 = 0.3085

step 5

Therefore,

b = 0.5

Answer

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

b = 0.5

Key Concept

Standard Normal Distribution and Probability Calculation

Explanation

The probability \operatorname{Pr}(X > 6) is

0.5

because

6

is the mean of

X

. For \operatorname{Pr}(X > 7) = \operatorname{Pr}(Z < b),

b

is found by converting

X

to the standard normal variable

Z

.

Generated Graph

Solution by Steps

step 1

Given the standard normal distribution, we have \operatorname{Pr}(Z<a)=A and \operatorname{Pr}(Z>b)=B, where b>a. We need to find \operatorname{Pr}(a<Z<b \mid Z<b)

step 2

Using the conditional probability formula, \operatorname{Pr}(a<Z<b \mid Z<b) = \frac{\operatorname{Pr}(a<Z<b \cap Z<b)}{\operatorname{Pr}(Z<b)}

step 3

Since a<Z<b is a subset of Z<b, \operatorname{Pr}(a<Z<b \cap Z<b) = \operatorname{Pr}(a<Z<b)

step 4

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

step 5

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

step 6

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

[Question 1] Answer

B

Key Concept

Conditional Probability

Explanation

Conditional probability is the probability of an event occurring given that another event has already occurred. In this case, we used the formula

\operatorname{Pr}(A \mid B) = \frac{\operatorname{Pr}(A \cap B)}{\operatorname{Pr}(B)}

.

Solution by Steps

step 1

Given a normal distribution

X

with mean

\mu = 60

and standard deviation

\sigma = 7

, we need to find the probability that

X

is less than 46

step 2

Convert

X

to the standard normal variable

Z

using

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

step 3

For

X = 46

,

Z = \frac{46 - 60}{7} = \frac{-14}{7} = -2

step 4

We need to find \operatorname{Pr}(Z < -2)

step 5

Using standard normal distribution tables or a calculator, \operatorname{Pr}(Z < -2) \approx 0.0228

[Question 2] Answer

A

Key Concept

Standard Normal Distribution

Explanation

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. Converting a normal variable to a standard normal variable allows us to use standard normal distribution tables to find probabilities.

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