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AI HL · 13.2

Probability Distributions

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13.2.9

Inverse normal calculations and normal practice

13.2.9

Inverse normal calculations and normal practice

Target: I can use inverse normal to find the value with a given area below it, and convert an upper-tail area to a lower-tail area first. (+6 more)

By the end of this section

Key points

Inverse normal

  • Inverse normal gives the value k when you know the area to the left of k. Enter the area, \(\mu\) and \(\sigma\).
  • For an upper tail, \(P(X > k) = p\), use the area \(1 - p\).
  • If a bound is given, as in \(P(a < X < k) = p\), add the area below a first: \(P(X < k) = P(X < a) + p\).
  • Sketch the curve with the area shaded. A top 10% gives a value above the mean; a bottom 10% gives a value below it.
  • If \(\mu\) or \(\sigma\) is unknown, write the probability with the unknown in the calculator's normal cdf, then solve for it. Check by putting the answer back.

Quartiles, percentiles and central intervals

  • The median of a normal distribution is the mean.
  • The lower quartile has area 0.25 to its left and the upper quartile has area 0.75. The IQR is \(Q_3 - Q_1\).
  • A box plot of a normal distribution is symmetric: the median line is in the middle of the box.
  • A central interval holds a percentage in the middle, such as 90%. Half of what is left goes in each tail: for 90%, the areas to the left are 0.05 and 0.95. The two ends are the same distance from the mean.

Using normal probabilities

  • Expected frequency = probability × sample size. Keep the unrounded probability for the multiplication.
  • Expected income: multiply each category's expected number by its price and add.
  • A normal probability can be the p of a binomial: \(Y \sim B(n, p)\). State that the trials are independent and p is constant.

When a normal model is poor

  • If the mean is less than 2 standard deviations above a limit such as 0, the model gives a noticeable probability to impossible values.
  • A negative result from a model of time, mass or length shows the model is only approximate. Say so.

Finding the mean and the standard deviation

  • Two probabilities give two equations. Find the z-value for each with inverse normal using \(\mu = 0\) and \(\sigma = 1\). Convert an upper tail first.
  • A value x has \(z = \frac{x - \mu}{\sigma}\), so \(x = \mu + z\sigma\). Write one equation for each value and solve the pair on the calculator.
  • Check: the probabilities from your \(\mu\) and \(\sigma\) must match the question.

On the calculator

Inverse normal

  • Enter the area to the left of the value, \(\mu\) and \(\sigma\).
  • The calculator gives the value x with that area below it.

Screenshot and video to come

Upper tail and between

  • For an area above, enter \(1 - p\).
  • For \(P(a < X < k) = p\), find \(P(X < a)\) with normal cdf, add p, and use that area.

Screenshot and video to come

Quartiles

  • Use inverse normal with area 0.25 for Q1 and 0.75 for Q3.
  • The median is the mean.

Screenshot and video to come

Unknown mean or standard deviation

  • Put the unknown in place of \(\mu\) or \(\sigma\) in the normal cdf, set it equal to the given probability, and solve for it.
  • Check by finding the probability again with your answer.

Screenshot and video to come

Explore: the normal distribution

A probability is an area under the curve. Drag the top of the curve to move the mean, the teal point on its side to change σ, and the orange points under the axis to choose the area. Or type exact values. Say how big the area is, then press Show answer.

Drag with the mouse or a finger, or select a point (Tab) and use the arrow keys. The axis fits itself to the curve when you let go.

Drag a point to change the shaded area.

  1. Drag the top of the curve, then the teal point. Which one moves the curve, and which one changes its shape? Why does a wider curve have to be lower?
  2. Press "Mark μ ± σ, 2σ, 3σ". Drag the orange points to μ − σ and μ + σ, then 2σ and 3σ. Say each area before you show it. Does it change when you change μ or σ?
  3. X ~ N(250, 16). What do you type for σ? Find P(X < 246).
  4. In "Find a value", choose Above and an area of 0.05. Is the value above or below the mean? Why does the calculator need the area 0.95?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Inverse normal, lower tail

The mass of a parcel is normally distributed with mean 14.2 kg and standard deviation 2.1 kg. The lightest 15% of parcels are sent by a cheaper service. Find the greatest mass of a parcel sent by the cheaper service.

4.26.38.410.512.614.716.818.92123.1Mass (kg)
Working
  1. Write the probability

    \(X \sim N(14.2, 2.1^2)\). \(P(X < k) = 0.15\)

    The lightest 15% are the lower tail. The area to the left of k is 0.15.

  2. Inverse normal

    From the calculator: inverse normal with area = 0.15, μ = 14.2, σ = 2.1

    Inverse normal takes the area to the left and gives the value.

  3. Answer

    \(k = 12.0\) kg

    Parcels lighter than this are about 15% of all parcels.

Video walkthrough coming
Inverse normal, upper tail

The daily takings of a market stall, in euros, are normally distributed with mean 300 and standard deviation 25. The stall owner says that on 5% of days the takings are above a value m. Find m.

200225250275300325350375400Takings (euros)
Working
  1. Convert to a lower tail

    \(P(X > m) = 0.05\), so \(P(X < m) = 0.95\)

    Inverse normal always works with the area to the left.

  2. Inverse normal

    From the calculator: inverse normal with area = 0.95, μ = 300, σ = 25

    The answer is above the mean, as it should be for a top 5%.

  3. Answer

    \(m = 341\)

    Check with a sketch: the shaded upper tail starts above the mean.

Video walkthrough coming
A bound is given

\(X \sim N(50, 8^2)\) Find the value of k such that \(P(45 < X < k) = 0.4\).

Working
  1. Area below the lower bound

    From the calculator: normal cdf with lower = −10^99, upper = 45, μ = 50, σ = 8: 0.2660

    P(X < 45) is the part to the left of 45.

  2. Area up to k

    \(P(X < k) = 0.2660 + 0.4 = 0.6660\)

    The area to the left of k is the area below 45 plus the 0.4 between 45 and k.

  3. Inverse normal

    From the calculator: inverse normal with area = 0.666, μ = 50, σ = 8: k = 53.4

    Use the full area, not 0.4. Keep the unrounded value in the calculator.

Video walkthrough coming
A symmetric interval

The lengths of a species of fish are normally distributed with mean 42 cm and standard deviation 5 cm. The middle 90% of lengths are called typical. Find the shortest and the longest typical length.

30405060Length (cm)
Working
  1. Share the tails

    Middle 90%, so 10% is left: 5% in each tail

    The curve is symmetric, so the two tails are equal.

  2. Upper end

    P(X < b) = 0.95. From the calculator: inverse normal with area = 0.95, μ = 42, σ = 5: b = 50.2

    The area to the left of the upper end is 0.95.

  3. Lower end

    P(X < a) = 0.05. From the calculator: inverse normal with area = 0.05, μ = 42, σ = 5: a = 33.8

    Check: the lower end is 8.22 below the mean and the upper end is 8.22 above it. The ends are equally far from the mean.

  4. Answer

    Typical lengths are from 33.8 cm to 50.2 cm

    Give both ends.

Video walkthrough coming
An unknown standard deviation

The mass of a bag of flour is normally distributed with mean 1 010 g. 4% of bags have a mass below 1 000 g. Find the standard deviation.

98099010001010102010301040Mass (g)
Working
  1. Write the condition

    X ~ N(1010, σ²). P(X < 1000) = 0.04

    σ is the unknown. 1 000 is below the mean, so the shaded tail is on the left.

  2. Solve for σ

    From the calculator: solve normal cdf with lower = −10^99, upper = 1000, μ = 1010, σ = s equal to 0.04, for s: s = 5.71

    Give the solver a starting value such as 5 if it asks.

  3. Check

    From the calculator: normal cdf with lower = −10^99, upper = 1000, μ = 1010, σ = 5.71…: 0.0400

    Putting the answer back gives the probability in the question.

Video walkthrough coming
Both unknown

The journey time, T minutes, of a bus is normally distributed. 20% of journeys take less than 18 minutes and 10% take more than 30 minutes. Find the mean and the standard deviation of T.

Working
  1. Two z-values

    P(T < 18) = 0.2: from the calculator: inverse normal with area = 0.2, μ = 0, σ = 1: z = -0.8416. P(T > 30) = 0.1, so P(T < 30) = 0.9: area = 0.9: z = 1.2816

    Standard normal means μ = 0, σ = 1. Convert the upper tail first.

  2. Two equations

    (18 − μ) ÷ σ = -0.8416 and (30 − μ) ÷ σ = 1.2816, so μ − 0.8416σ = 18 and μ + 1.2816σ = 30

    Each value is x = μ + zσ.

  3. Solve the pair

    From the calculator: simultaneous equations: σ = 5.65, μ = 22.8

    Keep the unrounded values for the check.

  4. Check

    From the calculator: normal cdf with μ = 22.757, σ = 5.652: P(T < 18) = 0.2000, P(T > 30) = 0.1000

    Both match the question.

Video walkthrough coming
Quartiles and a box plot

The heights of 300 plants are normally distributed with mean 64 cm and standard deviation 5 cm. Find the interquartile range, then draw a box plot using the shortest plant, 50 cm, and the tallest plant, 78 cm.

Working
  1. Lower quartile

    From the calculator: inverse normal with area = 0.25, μ = 64, σ = 5: Q1 = 60.6

    A quarter of the values are below the lower quartile.

  2. Upper quartile

    From the calculator: inverse normal with area = 0.75, μ = 64, σ = 5: Q3 = 67.4

    Three quarters are below the upper quartile.

  3. IQR

    IQR = Q3 − Q1 = 6.74

    Subtract the unrounded quartiles. The median of a normal distribution is the mean: 64.

  4. Plants455055606570758085Height (cm)
    Box plot

    Five numbers: 50, 60.6, 64, 67.4, 78

    The box runs from Q1 to Q3 with the median line at the mean, so the plot is symmetric.

Video walkthrough coming
Expected frequency

The lifetime of a kettle element is normally distributed with mean 520 days and standard deviation 30 days. A shop tests 1 200 elements. Find the expected number that last more than 580 days.

Working
  1. Probability

    From the calculator: normal cdf with lower = 580, upper = 10^99, μ = 520, σ = 30: 0.02275

    First the probability for one element.

  2. Expected frequency

    \(1 200 \times 0.02275 = 27.3\)

    Expected frequency = probability × sample size. Use the unrounded probability.

  3. Answer

    About 27 elements

    Round to a whole number of elements.

Video walkthrough coming
Normal, then binomial

The life of a battery is normally distributed with mean 9 hours and standard deviation 0.6 hours. A battery is faulty if it lasts less than 8 hours. A pack has 12 batteries. Find the probability that exactly 2 batteries in a pack are faulty.

Working
  1. Probability a battery is faulty

    From the calculator: normal cdf with lower = −10^99, upper = 8, μ = 9, σ = 0.6: p = 0.04779

    A normal probability becomes the p of a binomial.

  2. The binomial

    \(Y\) = number faulty, \(Y \sim B(12, 0.04779)\)

    Each battery is faulty or not, with the same p, independently.

  3. Exactly 2

    From the calculator: binomial pdf with n = 12, p = 0.04779, x = 2: 0.0924

    Assume batteries are independent and each has the same probability of being faulty.

Video walkthrough coming