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Answer #1

1.)

We are dealing with the standard normal distribution, with parameters:

μ=0, σ=1

We need to find a z-score z so that the corresponding cumulative normal probability is equal to 0.9718.

Mathematically, z is such that:

\Pr(Z \le z) = 0.9718

The corresponding z score so that the cumulative standard normal probability distribution is 0.9718 is

z_c = 1.9079

This value of z_c = 1.9079 can be found with a normal distribution table.

hence z = 1.9079

2)

The following information has been provided:

μ=100, σ=15

We need to compute Pr (75≤X≤115).

The corresponding z-values needed to be computed are:

Z_1 = \frac{X_1 - \mu}{\sigma}= \frac{ 75 - 100}{ 15} = -1.6667

Z_2 = \frac{X_2 - \mu}{\sigma} = \frac{ 115 - 100}{ 15}= 1

Therefore, we get:

\Pr(75 \leq X \leq 115) = \Pr\left(\frac{ 75 - 100}{ 15} \le Z \le \frac{ 115 - 100}{ 15}\right)= \Pr(-1.6667 \le Z \le 1)

= \Pr(Z \le 1) - \Pr(Z \le -1.6667) = 0.8413 - 0.0478 = 0.7936

3)

We know that X is normally distributed, with parameters:

μ=100, σ=15

Pr[X >= x] = 0.9 which is same as Pr[ X < x ] = 0.1

We need to find a score x so that the corresponding cumulative normal probability is equal to 0.1.

Mathematically, x is such that:

\Pr(X \le x) = 0.1

The corresponding z score so that the cumulative standard normal probability distribution is 0.1 is

z_c = -1.2816

This value of z_c = 1.2816 with a normal distribution table.

Hence, the X score associated with the 0.1 cumulative probability is

x = \mu + z_c \times \sigma = 100 + (-1.2816 \times 15) = 100 - 19.224 = 80.776

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