Question

suppose x is a normally distributed random variable with mean μ = 0 and standard deviation σ = 1. find the probability that x = -2.34. 9.62% 49.04% 99.04% 1.07%

Explanation:

Step1: Identify Distribution Type

The random variable \( x \) is normally distributed with \( \mu = 0 \) and \( \sigma = 1 \), so it's a standard normal distribution (\( Z \)-distribution), where \( Z = x \) (since \( Z=\frac{x - \mu}{\sigma}=\frac{x - 0}{1}=x \)). We need to find \( P(x \leq - 2.34) \) (assuming the problem means \( x \leq - 2.34 \); if it's \( x = - 2.34 \), the probability is 0, but the options suggest cumulative probability).

Step2: Use Z - Table

For a standard normal distribution, the Z - table gives the cumulative probability \( P(Z \leq z) \) for a given \( z \)-score. For \( z=-2.34 \), we look up the value in the Z - table. The Z - table value for \( z = - 2.34 \) (or equivalently, we can use the symmetry: \( P(Z \leq -z)=1 - P(Z \leq z) \), but directly looking up \( z=-2.34 \)):

The Z - table for \( z=-2.34 \): The row for \( - 2.3 \) and column for \( 0.04 \). The value is \( 0.0096 \) or \( 0.96\% \)? Wait, no, wait. Wait, maybe the problem is \( P(x \geq - 2.34) \)? Wait, no, let's re - check. Wait, the standard normal table: for \( z = - 2.34 \), the cumulative probability \( P(Z \leq - 2.34) \) is \( 0.0096 \) (or \( 0.96\% \))? But the options have \( 1.07\% \)? Wait, maybe a miscalculation. Wait, actually, the Z - table value for \( z=-2.34 \) is:

Looking at the Z - table: The left - tailed probability for \( z=-2.34 \) is calculated as follows. The Z - table gives \( P(Z \leq - 2.34)=0.0096 \approx 0.96\% \), but the closest option is \( 1.07\% \)? Wait, no, maybe the problem is \( P(x \geq - 2.34) \). Let's use symmetry: \( P(Z \geq - 2.34)=1 - P(Z \leq - 2.34) \). \( P(Z \leq - 2.34)=0.0096 \), so \( 1 - 0.0096 = 0.9904=99.04\% \)? No, that's one of the options. Wait, maybe the problem was \( P(x \geq - 2.34) \). Let's re - examine the options. The options are \( 9.62\% \), \( 49.04\% \), \( 99.04\% \), \( 1.07\% \).

Wait, if we consider \( P(-2.34\leq x\leq0) \), since the mean is 0, the area from \( - 2.34 \) to 0 is \( 0.5 - P(x \leq - 2.34) \). \( P(x \leq - 2.34)=0.0096 \), so \( 0.5 - 0.0096 = 0.4904 = 49.04\% \)? No, that's another option. Wait, no, let's start over.

The standard normal distribution has mean 0 and standard deviation 1. The total area under the curve is 1. The area to the left of \( z = 0 \) is 0.5, and to the right is 0.5.

For \( z=-2.34 \), the cumulative probability (area to the left of \( z=-2.34 \)) is found from the Z - table. The Z - table value for \( z=-2.34 \) is:

The Z - table: For \( z=-2.34 \), we look at the row for \( - 2.3 \) and column for \( 0.04 \). The value is \( 0.0096 \) (or \( 0.96\% \)). But the option \( 1.07\% \) is close (maybe due to rounding differences in the table). Wait, but if we consider \( P(x \geq - 2.34) \), that is \( 1 - P(x \leq - 2.34)=1 - 0.0096 = 0.9904 = 99.04\% \), which is one of the options (the third option: \( 99.04\% \)).

Wait, maybe the problem was to find \( P(x \geq - 2.34) \). Let's confirm:

In standard normal distribution, \( P(Z \geq - 2.34)=1 - P(Z \leq - 2.34) \). From Z - table, \( P(Z \leq - 2.34)=0.0096 \), so \( 1 - 0.0096 = 0.9904 = 99.04\% \), which matches the third option.

Answer:

99.04% (the option with 99.04%)