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

We need to find d where P(X < d) = 0.1151. First, we find the corresponding z-value from the z-table:

Z = -1.19 (approximately)

Now, we calculate d:

$d = ext{mean} + (z imes ext{standard deviation}) = 30 + (-1.19 imes 5) = 30 - 5.95 = 24.05.$

For P(X > e) = 0.1151, we have:

$P(X < e) = 1 - P(X > e) = 1 - 0.1151 = 0.8849.$

Looking up the corresponding z-value for 0.8849 in the z-table gives us:

Z = 1.2 (approximately)

Now, we find e:

$e = ext{mean} + (z imes ext{standard deviation}) = 30 + (1.2 imes 5) = 30 + 6 = 36.$

To find P(d < X < e), we can calculate:

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

From earlier calculations:

• P(X < e) = 0.8849
• P(X < d) = 0.1151

Thus, calculating gives us:

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