The document provides examples of solving probability problems using the normal distribution. It shows how to find the probability that a random variable falls below, above, or between given values using the standard normal distribution. The key steps are standardizing the random variable by subtracting the mean and dividing by the standard deviation, then looking up the standardized value in a normal distribution table. Examples calculate probabilities and estimates for situations like bulb lifetimes and milk delivery times.

2. Using Standard Normal Variable
Example-10:
Given a normal distribution with µ = 40 and  = 6 , find
(a). the area below 32,
(b). the area above 27,
(c). the area between 42 and 51.
Solution :
Given, µ = 40 and  = 6
X ̴ N(40,62
)
You need to find p(x < 32).To be able to use the standard normal
table, standardize the x variable by subtracting the mean 40 then
dividing by the standard deviation, 6. Appling this to both sides of the
inequality x<32.
(a). 𝑃(𝑥 < 32) = 𝑃 [
𝑥−𝜇
𝛿
<
32−𝜇
𝛿
]
= 𝑃 [𝑍 <
32−40
6
]
= 𝑃[ 𝑍 < −1.33]
= 0.0918
(b). 𝑃(𝑥 > 27) = 𝑃 [
𝑥−𝜇
𝛿
>
27−𝜇
𝛿
]
= 𝑃 [𝑍 >
27−40
6
]
= 𝑃[ 𝑍 > −2.17]
= 1 - 𝑃[ 𝑍 < −2.17]
= 1 − 0.0150 = 0.9850
32 40
27 40

3. (c). 𝑃(42 < 𝑥 < 51) = 𝑃 [
42−𝜇
𝛿
<
𝑥−𝜇
𝛿
<
51−𝜇
𝛿
]
= 𝑃 [
42−40
6
< 𝑍 <
51−40
6
]
= 𝑃[0.33 < 𝑍 < 1.83]
= 𝑃[0 < 𝑍 < 1.83] − 𝑃[0 < 𝑍 < 0.33]
= 0.4664 − 0.1293
= 0.3371
Example-11:
The life of a certain make of saver bulb is known to be normally
distributed with a mean life of 2000 hours and a standard deviation of
120 hours. In a batch of 5000 saver bulbs, estimate the number of
saver bulbs that the life of such a bulb will be
(a) Less than 1950 hour`s
(b) Greater than 2150 hour`s
(c) Less than 1850 hours and more than 2090 hours.
Solution :
(a). 𝑃(𝑥 < 1950) = 𝑃 [
𝑥−𝜇
𝛿
<
1950−𝜇
𝛿
]
= 𝑃 [𝑍 <
1950−2000
120
]
= 𝑃[ 𝑍 < −0.42]
= 0.3372
Number of saver bulb life less than 1950 hours are
0.3372×5000 = 1686 bulbs
40 42 51
1950 2000

4. (b). 𝑃(𝑥 > 2150) = 𝑃 [
𝑥−𝜇
𝛿
>
2150−𝜇
𝛿
]
= 𝑃 [𝑍 >
2150−2000
120
]
= 𝑃[ 𝑍 > 1.25]
= 1 - 𝑃[ 𝑍 < 1.25]
= 1 − 0.8944
= 0.1056
Number of saver bulb life greater than 2150 hours are
0.1056×5000 = 528 bulbs
(c). 𝑃(1850 < 𝑥 < 2090) = 𝑃 [
1850−𝜇
𝛿
<
𝑥−𝜇
𝛿
<
2090−𝜇
𝛿
]
= 𝑃 [
1850−2000
120
< 𝑍 <
2090−2000
120
]
= 𝑃[−1.25 < 𝑍 < 0.75]
= 𝑃[0 < 𝑍 < 0.75] − 𝑃[−1.25 < 𝑍 < 0]
= 0.7734 − 0.1056
= 0.6678
Number of saver bulb life between 1850 hours to 2090 hours are
0.6678×5000 = 3339 bulbs
2000 2150
1850 2000 2090

5. Using the Table in Reverse for any normal variable x
Example-12:
The time taken by the milkman to deliver to the high street is
normally distributed with a mean of 12 minutes and a standard
deviation of 2 minutes. He delivers milk every day.
Estimate the time of day during the year.
(a) Longer than 0.0062 . (b) Less than 0.1587.
Solution:
X is the time to delivers milk every day.
X ̴ N(12,22
)
(a) Given p(x > k) = 0.0062
Standardising
𝑃 [
𝑥−𝜇
𝛿
>
𝑘−𝜇
𝛿
] = 0.0062
𝑃 [𝑧 >
𝑘−12
2
] = 0.0062
Since
𝑘−12
2
= 2.50(The value of z at 0.0062 from
normal table is 2.50)
k = 2.50×2+12 = 17 minute
(b) Given p(x < k) = 0.1587
Standardising
𝑃 [
𝑥−𝜇
𝛿
<
𝑘−𝜇
𝛿
] = 0.1587
𝑃 [𝑧 <
𝑘−12
2
] = 0.1587
Since
𝑘−12
2
= −1.0(The value of z at 0.1587 from
normal table is -1.0)
k = (-1.0)×2+12 = 10 minute
12 k
0.0062
K 12
0.1587