The normal distribution Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level The normal distribution questions. See exactly how to solve problems on normal-distribution, probability, standardising, inverse-normal.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The random variable X∼N(50, 82)X\sim N(50,\ 8^{2}). Find P(X<58)P(X<58), giving your answer to 4 decimal places.

Worked solution

  1. Write down the distribution and required probability

    X∼N(50, 82),P(X<58)X\sim N(50,\ 8^{2}),\quad P(X<58)

    We are given a normal distribution and asked for a probability.

  2. Standardise and use the standard normal

    Z=X−μσ=58−508=1; P(X<58)=P(Z<1)=0.8413Z=\frac{X-\mu}{\sigma}=\frac{58-50}{8}=1;\ P(X<58)=P(Z<1)=0.8413

    Convert to Z, then read the probability from the standard normal.

  3. State the probability

    P(X<58)≈0.8413P(X<58)\approx 0.8413

    This is the required probability to 4 decimal places.

Answer
0.84130.8413
Question 2
2 markseasy
The random variable X∼N(100, 152)X\sim N(100,\ 15^{2}). Find P(X<85)P(X<85), giving your answer to 4 decimal places.

Worked solution

  1. Write down the distribution and required probability

    X∼N(100, 152),P(X<85)X\sim N(100,\ 15^{2}),\quad P(X<85)

    We are given a normal distribution and asked for a probability.

  2. Standardise and use the standard normal

    Z=X−μσ=85−10015=−1; P(X<85)=P(Z<−1)=0.1587Z=\frac{X-\mu}{\sigma}=\frac{85-100}{15}=-1;\ P(X<85)=P(Z<-1)=0.1587

    Convert to Z, then read the probability from the standard normal.

  3. State the probability

    P(X<85)≈0.1587P(X<85)\approx 0.1587

    This is the required probability to 4 decimal places.

Answer
0.15870.1587
Question 3
2 markseasy
The random variable X∼N(20, 42)X\sim N(20,\ 4^{2}). Find P(X>26)P(X>26), giving your answer to 4 decimal places.

Worked solution

  1. Write down the distribution and required probability

    X∼N(20, 42),P(X>26)X\sim N(20,\ 4^{2}),\quad P(X>26)

    We are given a normal distribution and asked for a probability.

  2. Standardise and use the standard normal

    Z=X−μσ=26−204=1.5; P(X>26)=P(Z>1.5)=1−0.9332=0.0668Z=\frac{X-\mu}{\sigma}=\frac{26-20}{4}=1.5;\ P(X>26)=P(Z>1.5)=1-0.9332=0.0668

    Convert to Z, then read the probability from the standard normal.

  3. State the probability

    P(X>26)≈0.0668P(X>26)\approx 0.0668

    This is the required probability to 4 decimal places.

Answer
0.06680.0668
Question 4
2 markseasy
The random variable X∼N(70, 102)X\sim N(70,\ 10^{2}). Find P(X>55)P(X>55), giving your answer to 4 decimal places.

Worked solution

  1. Write down the distribution and required probability

    X∼N(70, 102),P(X>55)X\sim N(70,\ 10^{2}),\quad P(X>55)

    We are given a normal distribution and asked for a probability.

  2. Standardise and use the standard normal

    Z=X−μσ=55−7010=−1.5; P(X>55)=P(Z>−1.5)=1−0.0668=0.9332Z=\frac{X-\mu}{\sigma}=\frac{55-70}{10}=-1.5;\ P(X>55)=P(Z>-1.5)=1-0.0668=0.9332

    Convert to Z, then read the probability from the standard normal.

  3. State the probability

    P(X>55)≈0.9332P(X>55)\approx 0.9332

    This is the required probability to 4 decimal places.

Answer
0.93320.9332
Question 5
2 markseasy
The random variable X∼N(30, 52)X\sim N(30,\ 5^{2}). Find P(X<37)P(X<37), giving your answer to 4 decimal places.

Worked solution

  1. Write down the distribution and required probability

    X∼N(30, 52),P(X<37)X\sim N(30,\ 5^{2}),\quad P(X<37)

    We are given a normal distribution and asked for a probability.

  2. Standardise and use the standard normal

    Z=X−μσ=37−305=1.4; P(X<37)=P(Z<1.4)=0.9192Z=\frac{X-\mu}{\sigma}=\frac{37-30}{5}=1.4;\ P(X<37)=P(Z<1.4)=0.9192

    Convert to Z, then read the probability from the standard normal.

  3. State the probability

    P(X<37)≈0.9192P(X<37)\approx 0.9192

    This is the required probability to 4 decimal places.

Answer
0.91920.9192

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