The weight, X grams, of soup put in a tin by machine A is normally distributed with a mean of 160 g and a standard deviation of 5 g - Edexcel - A-Level Maths: Statistics - Question 8 - 2011 - Paper 1

To find the value of w such that P(X < w) = 0.01, we need to find the Z-score corresponding to this probability:

Using a standard normal distribution table, we find:

$Z = -2.3263$

Now, we revert to the original variable X using:

$w = 160 + Z * 5$

Substituting the Z-value:

$w = 160 + (-2.3263) * 5 = 160 - 11.6315 \approx 148.37$

Thus, w is approximately 148.37.

From the given probabilities, we can use the Z-score formulas:

1. For P(Y < 160) = 0.99:

   • The Z-score for 0.99 is approximately 2.33: $2.33 = \frac{160 - \mu}{\sigma}$ Therefore, we have: $160 - \mu = 2.33\sigma \Rightarrow \mu = 160 - 2.33\sigma$

2. For P(Y > 152) = 0.90:

   • This means P(Y < 152) = 0.10, with a corresponding Z-score of approximately -1.28: $-1.28 = \frac{152 - \mu}{\sigma}$ Therefore: $152 - \mu = -1.28\sigma \Rightarrow \mu = 152 + 1.28\sigma$

Now we have a system of two equations:

1. ( \mu = 160 - 2.33\sigma )
2. ( \mu = 152 + 1.28\sigma )

Equating the two expressions for μ:

$160 - 2.33\sigma = 152 + 1.28\sigma$

Solving for σ:

$8 = 3.61\sigma \Rightarrow \sigma \approx 2.21$

Substituting back to find μ:

$\mu = 160 - 2.33(2.21) \approx 154.85$

Thus, the values are approximately:
μ ≈ 154.85 and σ ≈ 2.21.