The random variable X has a normal distribution with mean 30 and standard deviation 5 - Edexcel - A-Level Maths Statistics - Question 6 - 2009 - Paper 1

To find d, we need to identify the z-value corresponding to the cumulative probability of 0.1151. Using the standard normal distribution, we find:

$$-1.1151 \approx z$$

Using the transformation formula for d:

$$d = \mu + z \cdot \sigma = 30 + (-1.1151) \cdot 5 = 24.425$$

Thus, the value of d is approximately 24.425.

To find e, we recognize that P(X > e) = 0.1151 implies:

$$P(X < e) = 1 - P(X > e) = 0.8849$$

Now we find the corresponding z-value:

$$z \approx 1.2$$

Using the transformation formula for e:

$$e = \mu + z \cdot \sigma = 30 + (1.2) \cdot 5 = 36$$

Thus, the value of e is 36.

To find P(d < X < e), we can express it as:

$$P(d < X < e) = P(X < e) - P(X < d)$$

From previous calculations, we know:

$$P(X < e) = 0.8849\, \text{and}\, P(X < d) \approx 0.1151$$

Substituting these values gives:

$$P(d < X < e) = 0.8849 - 0.1151 = 0.7698$$

Thus, P(d < X < e) is approximately 0.7698.