Question

A population of values has a normal distribution with μ = 155.4 and σ = 49.5 . You intend to draw a random sample of size n = 246 . Find the probability that a single randomly selected value is between 158.6 and 159.2. P(158.6 < X < 159.2) = .0048 Correct Find the probability that a sample of size n = 246 is randomly selected with a mean between 158.6 and 159.2. P(158.6 < M < 159.2) = .0410 Correct

Answer 1

Answer:

(a) The probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.

(b) The probability that a sample mean is between 158.6 and 159.2 is 0.0411.

Step-by-step explanation:

Let the random variable X follow a Normal distribution with parameters μ = 155.4 and σ = 49.5.

(a)

Compute the probability that a single randomly selected value lies between 158.6 and 159.2 as follows:

$P(158.6 < X$

*Use a standard normal table.

Thus, the probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.

(b)

A sample of n = 246 is selected.

Compute the probability that a sample mean is between 158.6 and 159.2 as follows:

$P(158.6 < \bar X$

*Use a standard normal table.

Thus, the probability that a sample mean is between 158.6 and 159.2 is 0.0411.

Answer 2

Answer:

(a) P(158.6 < X < 159.2) = 0.0048

(b) P(158.6 < M < 159.2) = 0.041

Step-by-step explanation:

We are given that a population of values has a normal distribution with μ = 155.4 and σ = 49.5.

(a) Let X = a single randomly selected value

So, X ~ N($\mu=155.4,\sigma^{2} = 49.5^{2}$)

The z-score probability distribution for single selected value is given by;

                Z = $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean = 155.4

            $\sigma$ = standard deviation = 149.5

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that a single randomly selected value is between 158.6 and 159.2 is given by =  P(158.6 < X < 159.2) = P(X < 159.2) - P(X $\leq$ 158.6)

    P(X < 159.2) = P( $\frac{X-\mu}{\sigma}$ < $\frac{159.2-155.4}{49.5}$ ) = P(Z < 0.077) = 0.5307

    P(X $\leq$ 158.6) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{158.6-155.4}{49.5}$ ) = P(Z $\leq$ 0.065) = 0.5259                                    

The above probabilities is calculated by looking at the value of x = 0.077 and x = 0.065 in the z table which has an area of 0.5307 and 0.5259 respectively.

Therefore, P(158.6 < X < 159.2) = 0.5307 - 0.5259 = 0.0048

(b) Now we are given a sample size of 246.

Let M = sample mean

The z-score probability distribution for sample mean is given by;

                Z = $\frac{ M -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$  ~ N(0,1)

where, $\mu$ = population mean = 155.4

            $\sigma$ = standard deviation = 149.5

            n = sample size = 246

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the mean is between 158.6 and 159.2 is given by =  P(158.6 < M < 159.2) = P(M < 159.2) - P(M $\leq$ 158.6)

    P(M < 159.2) = P( $\frac{ M -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 159.2-155.4}{{\frac{49.5}{\sqrt{246} } }} }$ ) = P(Z < 1.20) = 0.885

    P(M $\leq$ 158.6) = P( $\frac{ M -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 158.6-155.4}{{\frac{49.5}{\sqrt{246} } }} }$ ) = P(Z $\leq$ 1.01) = 0.844                                      

The above probabilities is calculated by looking at the value of x = 1.20 and x = 1.01 in the z table which has an area of 0.885 and 0.844 respectively.

Therefore, P(158.6 < M < 159.2) = 0.885 - 0.844 = 0.041