AI HL · 13.4
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Exam question: sums of normals, a test for a mean and Type II error
13.4.21
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The mass, X grams, of a fish fillet from a fishmonger is normally distributed with mean 180 and standard deviation 9. (a) Find the probability that one fillet has a mass greater than 190 g. (b) Two fillets are chosen at random. Find the probability that both have a mass greater than 190 g.
Have a go first. Then open the working one line at a time.
From the calculator: normal cdf, lower bound 190, upper bound 1E99, mean 180, standard deviation 9: 0.133 P(X > 190) = 0.133
Normal cdf with lower bound 190 and no upper bound.
Two independent items, both above 190: multiply. P(both) = 0.1333² = 0.0178
The fillets are independent, so P(both) = P(first) × P(second).
The mass, X grams, of a fish fillet from a fishmonger is normally distributed with mean 180 and standard deviation 9. Two fillets are chosen at random. Find the probability that their total mass is greater than 385 g.
Have a go first. Then open the working one line at a time.
E(X₁ + X₂) = 180 + 180 = 360
Means add.
Var(X₁ + X₂) = 9² + 9² = 162, so standard deviation = 12.73
Variances add for independent variables. Standard deviations do not.
From the calculator: normal cdf, lower bound 385, upper bound 1E99, mean 360, standard deviation 12.73: 0.0248
T ~ N(360, 162).
A fishmonger changes the feed of his fish. He thinks the mean mass of a fillet has increased. The standard deviation stays 9 g and the old mean was 180 g. He takes a random sample of 25 fillets and tests at the 1% significance level. (a) State the null and alternative hypotheses. (b) Find the critical region for the sample mean.
Have a go first. Then open the working one line at a time.
H0: μ = 180, H1: μ > 180
H1 is one-tailed because he thinks the mean has increased. The value 180 must appear in both.
\(\bar X \sim N(180, 9^2/25)\), standard deviation 9/√25 = 1.80
Divide σ by √n for the sample mean.
From the calculator: inverse normal, area 0.99, mean 180, standard deviation 1.80: 184
1% of sample means lie above this value when H0 is true.
\(\bar X > 184\)
Write it as an inequality for the sample mean.
The new feed makes the mass of every fillet 3% greater than it would have been. The test above is carried out with the critical region \(\bar X > 184\), and the sample is of 25 fillets. (a) Find the mean and standard deviation of the new distribution of masses. (b) Find the probability of a Type II error. Use the unrounded critical value.
Have a go first. Then open the working one line at a time.
A 3% increase multiplies by 1.03: \(W = 1.03X\) E(W) = 1.03 × 180 = 185.4
A percentage increase multiplies the mean.
SD(W) = 1.03 × 9 = 9.27
The standard deviation is multiplied by 1.03 too. The variance is multiplied by 1.03².
\(\bar X \sim N(185.4, 9.27^2/25)\), standard deviation 1.85
Use the new distribution to find a Type II error.
Type II error: \(\bar X < 184.187\) From the calculator: normal cdf, lower bound −1E99, upper bound 184.187 (the unrounded critical value), mean 185.4, standard deviation 1.85: 0.257
Not rejecting H0 is the area below the critical value on the new curve.
P(Type II error) = 0.257
The test misses a 3% rise about a quarter of the time.