Question

P(x < 21 | μ = 23 and σ = 3) enter the probability of fewer than 21 outcomes if the mean is 23 and the standard deviation is 3 (b) P(x ≥ 66 | μ = 50 and σ = 9) enter the probability of 66 or more outcomes if the mean is 50 and the standard deviation is 9 (c) P(x > 47 | μ = 50 and σ = 5) enter the probability of more than 47 outcomes if the mean is 50 and the standard deviation is 5 (d) P(17 < x < 24 | μ = 21 and σ = 3) enter the probability of more than 17 and fewer than 24 outcomes if the mean is 21 and the standard deviation is 3 (e) P(x ≥ 95 | μ = 80 and σ = 1.82) enter the probability of 95 or more outcomes if the mean is 80 and the standard deviation is 1.82

Answer 1

Answer:

(a) The value of P (X < 21 | μ = 23 and σ = 3) is 0.2514.

(b) The value of P (X ≥ 66 | μ = 50 and σ = 9) is 0.0427.

(c) The value of P (X > 47 | μ = 50 and σ = 5) is 0.7258.

(d) The value of P (17 < X < 24 | μ = 21 and σ = 3) is 0.7495.

(e) The value of P (X ≥ 95 | μ = 80 and σ = 1.82) is 0.

Step-by-step explanation:

The random variable X is Normally distributed.

(a)

The mean and standard deviation are:

$\mu=23\\\sigma=3$

Compute the value of P (X < 21) as follows:

$P(X$

                  $=P(Z$

Thus, the value of P (X < 21 | μ = 23 and σ = 3) is 0.2514.

(b)

The mean and standard deviation are:

$\mu=50\\\sigma=9$

Compute the value of P (X ≥ 66) as follows:

Use continuity correction.

P (X ≥ 66) = P (X > 66 - 0.5)

                = P (X > 65.5)

                $=P(\frac{X-\mu}{\sigma}>\frac{65.5-50}{9})$

                $=P(Z>1.72)\\=1-P(Z$

Thus, the value of P (X ≥ 66 | μ = 50 and σ = 9) is 0.0427.

(c)

The mean and standard deviation are:

$\mu=50\\\sigma=5$

Compute the value of P (X > 47) as follows:

$P(X>47)=P(\frac{X-\mu}{\sigma}>\frac{47-50}{5})$

                 $=P(Z>-0.60)\\=P(Z$

Thus, the value of P (X > 47 | μ = 50 and σ = 5) is 0.7258.

(d)

The mean and standard deviation are:

$\mu=21\\\sigma=3$

Compute the value of P (17 < X < 24) as follows:

$P(17$

                          $=P(-1.33$

Thus, the value of P (17 < X < 24 | μ = 21 and σ = 3) is 0.7495.

(e)

The mean and standard deviation are:

$\mu=80\\\sigma=1.82$

Compute the value of P (X ≥ 95) as follows:

Use continuity correction:

P (X ≥ 95) = P (X > 95 - 0.5)

                = P (X > 94.5)

                $=P(\frac{X-\mu}{\sigma}>\frac{94.5-80}{1.82})$

                $=P(Z>7.97)\\=1-P(Z$

Thus, the value of P (X ≥ 95 | μ = 80 and σ = 1.82) is 0.