Name: Random Variables in Probability Theory
Uploaded: 2019-07-19
Duration: 8 min 57 s
Description: What do we mean by random variables? Watch this video to learn about random variables alongside continuous random variables and discrete random variables wit...

Question

X

has a normal distribution, such that

P (X > 147.0) = 0.0136

and

P (X ⩽ 59.0) = 0.0038

. Use this information to calculate the probability that

80.0 ⩽ X < 130.0

.

(Round answer up to $2$ decimal places)

Accepted Answer

Given: normally distributed variable is $P (X > 147.0) = 0.0136$ , where $μ$ is the mean and $σ$ is the standard deviation.

Given: $P (X \leq 59) = 0.0038$ , $P (X > 147) = 0.0136$

After standardising,

$\Rightarrow P (Z \leq z_{59}) = 0.0038$

$P (Z \leq z_{59}) < 0.5$ tells us that $z_{59}$ is negative.

Let $z_{59} = - a, P (Z \leq - a) = 0.0038$

We know that, if $z = - a$ , $P (Z \leq - a) = 1 - ϕ (a)$ .

Then, $1 - ϕ (a) = 0.0038$

$\Rightarrow ϕ (a) = 0.9962$

To find a value of $z$ , we use the tables in reverse and search for the $ϕ (z)$ .

$\Rightarrow a = 2.67 \Rightarrow z_{59} = - a = - 2.67$

For $P (X > 147) = 0.0136$ , $P (Z > z_{147}) = 0.0136$

$P (Z > z_{147}) < 0.5$ tells us that $z_{147}$ is positive.

If $z = b$ , $P (Z \geq b) = 1 - ϕ (b)$ .

$\Rightarrow 1 - ϕ (z_{147}) = 0.0136 \Rightarrow ϕ (z_{147}) = 0.9864 \Rightarrow z_{147} = 2.21$

We know that, standardised value for variable $X$ is $z = \frac{x - μ}{σ}$ .

When $x = 59$ , $z_{59} = - 2.67$

Now, $z_{59} = \frac{59 - μ}{σ}$

$\Rightarrow - 2.67 = \frac{59 - μ}{σ} \Rightarrow - 2.67 σ = 59 - μ \Rightarrow μ = 59 + 2.67 σ . . . . . . . . (i) e q .$

When $x = 147$ , $z_{147} = 2.21$

Now, $z_{147} = \frac{147 - μ}{σ}$

$\Rightarrow 2.21 = \frac{147 - μ}{σ} \Rightarrow 2.21 σ = 147 - (59 + 2.67 σ) \Rightarrow 4.88 σ = 88 \Rightarrow σ = 18.03$

From $(i) e q .$ , $μ = 59 + 2.67 \times 18.03$

$\Rightarrow μ = 107.1401 \approx 107.14$ .

Now, $P (80.0 ⩽ X < 130.0)$

Standardising score for $x = 80$ , $z = \frac{80 - 107.14}{18.03} = - 1.505$

Standardising score for $x = 130$ , $z = \frac{130 - 107.14}{18.03} = 1.268$

$\Rightarrow P (80.0 ⩽ X < 130.0) = P (- 1.505 ⩽ Z < 1.268) \Rightarrow P (80.0 ⩽ X < 130.0) = P (Z < 1.268) - P (Z < - 1.505) \Rightarrow P (80.0 ⩽ X < 130.0) = ϕ (1.268) - [1 - ϕ (1.505)] \Rightarrow P (80.0 ⩽ X < 130.0) = 0.8977 - 1 + 0.9338 \Rightarrow P (80.0 ⩽ X < 130.0) = 0.8315$

$X$ has a normal distribution, such that $P (X > 147.0) = 0.0136$ and $P (X ⩽ 59.0) = 0.0038$ . Use this information to calculate the probability that $80.0 ⩽ X < 130.0$ .
(Round answer up to $2$ decimal places)

Important Questions on The Normal Distribution

Learn and Prepare for any exam you want !

X has a normal distribution, such that P(X>147.0)=0.0136 and P(X⩽59.0)=0.0038. Use this information to calculate the probability that 80.0⩽X<130.0. (Round answer up to 2 decimal places)

Important Questions on The Normal Distribution

Learn and Prepare for any exam you want !

$X$ has a normal distribution, such that $P (X > 147.0) = 0.0136$ and $P (X ⩽ 59.0) = 0.0038$ . Use this information to calculate the probability that $80.0 ⩽ X < 130.0$ .
(Round answer up to $2$ decimal places)