Inverse normal and normal practice Problem set B One question at a time

The time a runner takes to complete a 200 m race is normally distributed with mean 25.4 s and standard deviation 1.2 s. The fastest 5% of runners reach the final. Find the slowest time that reaches the final.

• P(X < k) = 0.05
• From the calculator: inverse normal with area = 0.05, μ = 25.4, σ = 1.2
• k = 23.4

The monthly pay of workers in a factory is normally distributed with mean €2 400 and standard deviation €350. The top 12% of earners pay a higher rate of tax. Find the lowest pay that is taxed at the higher rate.

• P(X > k) = 0.12, so P(X < k) = 0.88
• From the calculator: inverse normal with area = 0.88, μ = 2 400, σ = 350
• k = 2810

\(X \sim N(45, 6^2)\). Find the value of k such that \(P(40 < X < k) = 0.45\).

• From the calculator: normal cdf with lower = −10^99, upper = 40, μ = 45, σ = 6: 0.2023
• \(P(X < k) = 0.2023 + 0.45 = 0.6523\)
• From the calculator: inverse normal with area = 0.6523, μ = 45, σ = 6
• k = 47.3

The lengths of rope cut by a machine are normally distributed with mean 2.40 m and standard deviation 0.06 m.

1. (a) Find the lower quartile.

   • From the calculator: inverse normal with area = 0.25, μ = 2.4, σ = 0.06
   • Q1 = 2.36

2. (b) Find the upper quartile.

   • From the calculator: inverse normal with area = 0.75, μ = 2.4, σ = 0.06
   • Q3 = 2.44

3. (c) Find the interquartile range.

   • IQR = Q3 − Q1 = 0.0809 (from the unrounded quartiles)

The lifetime of a smartphone battery is normally distributed with mean 820 days and standard deviation 90 days. Find the probability that a battery lasts less than 700 days.

• \(X \sim N(820, 90^2)\)
• From the calculator: normal cdf with lower = −10^99, upper = 700, μ = 820, σ = 90
• P = 0.0912

The length of a pencil after sharpening is normally distributed with mean 17.5 cm and standard deviation 0.8 cm. In a box of 1 200 pencils, find the expected number longer than 18.5 cm.

• From the calculator: normal cdf with lower = 18.5, upper = 10^99, μ = 17.5, σ = 0.8: 0.1056
• \(1 200 \times 0.1056 = 127\)
• About 127 pencils

The braking distance of a model of car from 50 km/h is normally distributed with mean 24.5 m and standard deviation 1.8 m. The middle 80% of distances are called usual. Find the shortest and longest usual distance.

• The middle 80% leaves 20%, so 10% in each tail
• From the calculator: inverse normal with area = 0.1, μ = 24.5, σ = 1.8: 22.2
• From the calculator: inverse normal with area = 0.9, μ = 24.5, σ = 1.8: 26.8
• The interval is 22.2 to 26.8 m

\(X \sim N(\mu, 4.5^2)\) and \(P(X < 62) = 0.15\). Find the value of \(\mu\).

• The unknown is μ. σ = 4.5
• From the calculator: solve normal cdf with lower = −10^99, upper = 62, μ = m, σ = 4.5 equal to 0.15, for m
• μ = 66.7

The time a student takes to complete a cross-country course is normally distributed with mean 52 minutes and standard deviation 6 minutes. 90% of students finish within t minutes. Find t.

• \(P(X < t) = 0.9\)
• From the calculator: inverse normal with area = 0.9, μ = 52, σ = 6
• t = 59.7 minutes

The mass of a peach is normally distributed with mean 150 g and standard deviation 20 g. A grower sorts 3 000 peaches. Small peaches (under 130 g) sell for €0.25 each, medium peaches (130 g to 170 g) for €0.40 each and large peaches (over 170 g) for €0.60 each. Find the expected income.

• P(small) = 0.159, P(medium) = 0.683, P(large) = 0.159 (calculator values, kept unrounded)
• Expected numbers: 476 small, 2048 medium, 476 large
• \(476 \times 0.25 + 2050 \times 0.40 + 476 \times 0.60\), using the unrounded numbers
• Expected income = €1 223.80

The distance, in kilometres, that a delivery driver travels to a customer is modelled by a normal distribution with mean 4 and standard deviation 2.2.

1. (a) Find the probability that the model gives a negative value.

   • From the calculator: normal cdf with lower = −10^99, upper = 0, μ = 4, σ = 2.2
   • P(X < 0) = 0.0345

2. (b) What does your answer tell you about the model?

   • A distance cannot be negative, but the model gives it probability 0.0345.
   • The mean is close to 0 relative to the spread, so the normal curve is a poor fit.

The mass of a bag from a packing machine is normally distributed with mean 1 015 g and standard deviation 12 g. A bag is underweight if its mass is less than 1 000 g. A sample of 15 bags is taken.

1. (a) Find the probability that one is below the limit given in the question.

   • From the calculator: normal cdf with lower = −10^99, upper = 1 000, μ = 1 015, σ = 12
   • p = 0.106

2. (b) Find the probability that at least 2 bags in the sample are underweight.

   • \(Y \sim B(15, 0.1056)\)
   • From the calculator: binomial cdf with n = 15, p = 0.1056, upper = 1: 0.5193
   • P(Y ≥ 2) = 1 − 0.5193 = 0.481

3. (c) State two assumptions you made in part (b).

   • The events are independent of each other.
   • The probability p is constant for every one.

A machine fills bottles with juice. The volume is normally distributed with mean 330 ml. 2% of bottles contain less than 325 ml.

1. (a) Find the standard deviation.

   • X ~ N(330, σ²). P(X < 325) = 0.02
   • From the calculator: solve normal cdf with lower = −10^99, upper = 325, μ = 330, σ = s equal to 0.02, for s
   • σ = 2.43

2. (b) Find the probability that a bottle holds more than 333 ml.

   • P(X > 333) = 1 − P(X < 333)
   • From the calculator: normal cdf with lower = 333, upper = 10^99, μ = 330, σ = 2.4346…: 0.109

The lengths of 500 trout are normally distributed with mean 32 cm and standard deviation 4 cm. The shortest trout is 19 cm and the longest 45 cm. Complete a box plot of the lengths.

1. (a) Find the median, lower quartile and upper quartile.

   • The median of a normal distribution is the mean: 32
   • From the calculator: inverse normal with area = 0.25, μ = 32, σ = 4: 29.3
   • From the calculator: inverse normal with area = 0.75, μ = 32, σ = 4: 34.7

2. (b) Write the five numbers for the box plot and describe its shape.

   • Five numbers: 19, 29.3, 32, 34.7, 45
   • Draw the box from Q1 to Q3, the median line at the mean, and whiskers to the lowest and highest values. The plot is symmetric.

The number of days that a parcel is delayed is modelled by a normal distribution with mean 4.0 and standard deviation 2.8.

1. (a) How many standard deviations is the mean above 0?

   • \(4 \div 2.8 = 1.43\)

2. (b) Explain why a normal distribution is not a suitable model.

   • 1.43 is less than 2.
   • From the calculator: normal cdf with lower = −10^99, upper = 0, μ = 4, σ = 2.8: 0.0766
   • The model gives a probability of 0.0766 to a negative delay, which cannot happen.

The life, L hours, of a type of battery is normally distributed. 15% of batteries last less than 18 hours and 25% last more than 30 hours.

1. (a) Find the mean and the standard deviation.

   • P(L < 18) = 0.15: inverse normal with area = 0.15, μ = 0, σ = 1: z = -1.0364
   • P(L > 30) = 0.25, so area = 0.75: z = 0.6745
   • μ − 1.0364σ = 18 and μ + 0.6745σ = 30
   • From the calculator: simultaneous equations: σ = 7.01, μ = 25.3

2. (b) Find the probability that a battery lasts more than 36 hours.

   • From the calculator: normal cdf with lower = 36, upper = 10^99, μ = 25.2693…, σ = 7.0138…: 0.063