Question

If mean =70 and Standard deviation is 16 i) P(38≤X≤46) ii) P(82≤X≤94) iii) P(62≤X≤86) Find the Probability values and Plot the graph with text. (Standard Normal distribution or General Normal Distribution.)

Answer 1

`i) \( P(38 \leq X \leq 46) \approx 0.197 \)`

`ii) \( P(82 \leq X \leq 94) \approx 0.155 \)`

`iii) \( P(62 \leq X \leq 86) \approx 0.707 \)`

Explanation:

To find these probabilities, we need to standardize the values using the z-score formula:
`\( Z = \frac{{X - \mu}}{{\sigma}} \)`, where
`\( \mu \)` is the mean and
`\( \sigma \)` is the standard deviation.

For the first case (i), standardizing 38 and 46:

`\[ Z_1 = \frac{{38 - 70}}{{16}} \approx -2 \]`

`\[ Z_2 = \frac{{46 - 70}}{{16}} \approx -1.5 \]`

Using a standard normal distribution table or calculator, the probability is
`\( P(-2 \leq Z \leq -1.5) \approx 0.197 \).`

For the second case (ii), standardizing 82 and 94:

`\[ Z_3 = \frac{{82 - 70}}{{16}} \approx 0.75 \]`

`\[ Z_4 = \frac{{94 - 70}}{{16}} \approx 1.5 \]`

The probability is
`\( P(0.75 \leq Z \leq 1.5) \approx 0.155 \).`

Finally, for the third case (iii), standardizing 62 and 86:

`\[ Z_5 = \frac{{62 - 70}}{{16}} \approx -0.5 \]`

`\[ Z_6 = \frac{{86 - 70}}{{16}} \approx 1 \]`

The probability is
`\( P(-0.5 \leq Z \leq 1) \approx 0.707 \).`

In conclusion, by standardizing the values and referring to the standard normal distribution, we find the probabilities for the given intervals. The probabilities represent the likelihood of a random variable falling within those specific ranges.