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 $w$ such that $P(X < w) = 0.01$, first we standardize: $P(X < w) = P\left(Z < \frac{w - \mu}{\sigma}\right)$

Setting this equal to 0.01: $\frac{w - 160}{5} = Z_{0.01}$

From standard normal distribution tables, we find that: $Z_{0.01} \approx -2.326$

Now, rearranging gives: $w - 160 = -2.326 \cdot 5$

Calculating: $w - 160 \approx -11.63$

Thus, $w \approx 148.37$

Hence, $w$ is approximately 148.37 grams.

We need to establish the values of $\mu$ and $\sigma$ based on the probabilities provided.

1. Finding $\mu$: From $P(Y < 160) = 0.99$, we standardize: $P\left(Z < \frac{160 - \mu}{\sigma}\right) = 0.99$ Therefore, using the Z-table: $\frac{160 - \mu}{\sigma} \approx 2.33$

2. Finding $\sigma$: From $P(Y > 152) = 0.90$, we have: $P(Y < 152) = 0.10$ Standardizing gives: $P\left(Z < \frac{152 - \mu}{\sigma}\right) = 0.10$ Therefore: $\frac{152 - \mu}{\sigma} \approx -1.281$

Now we have a system of equations:

1. $\frac{160 - \mu}{\sigma} = 2.33 \Rightarrow 160 - \mu = 2.33\sigma$
2. $\frac{152 - \mu}{\sigma} = -1.281 \Rightarrow 152 - \mu = -1.281\sigma$

Solving these:

1. $\mu = 160 - 2.33\sigma$
2. $\mu = 152 + 1.281\sigma$

Setting these equal: $160 - 2.33\sigma = 152 + 1.281\sigma$ $8 = 3.611\sigma$ $\sigma \approx 2.21$

Substituting back to find $\mu$: $\mu = 160 - 2.33(2.21) \approx 154.84$

Hence, the values are: $\mu \approx 154.84$ and $\sigma \approx 2.21$.