Question

A happiness survey with normally distributed scores with mean 5.5 and standard deviation 2.3 was administered to IT workers in healthcare. a) Find the probability that a randomly selected participant's response was greater than 5. 2- 5-5.9 2= -0.2174 0.4129 2.3 -0.4129 -0.5871 b) Find the probability that a randomly selected participant's response was between 4.5 and 6.5. 6.5-5.5 2-0.434783 22.30 = -0.434783 0.6604-0-5536 = 2.3 0.6664 0.3336 [0.3328 c) Find the probability that the mean of a sample of 16 selected participant's response was between 4.5 and 6.5.

Answer 1

Answer:

a) 58.71% probability that a randomly selected participant's response was greater than 5

b) 33.28% probability that a randomly selected participant's response was between 4.5 and 6.5.

c) 91.82% probability that the mean of a sample of 16 selected participant's response was between 4.5 and 6.5.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 5.5, \sigma = 2.3$

a) Find the probability that a randomly selected participant's response was greater than 5

This probability is 1 subtracted by the pvalue of Z when X = 5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 5.5}{2.3}$

$Z = -0.22$

$Z = -0.22$ has a pvalue of 0.4129.

1 - 0.4129 = 0.5871

58.71% probability that a randomly selected participant's response was greater than 5

b) Find the probability that a randomly selected participant's response was between 4.5 and 6.5.

This probability is the pvalue of Z when X = 6.5 subtracted by the pvalue of Z when X = 4.5. So

X = 6.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6.5 - 5.5}{2.3}$

$Z = 0.43$

$Z = 0.43$ has a pvalue of 0.6664.

X = 4.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.5 - 5.5}{2.3}$

$Z = -0.43$

$Z = -0.43$ has a pvalue of 0.3336

0.6664 - 0.3336 = 0.3328

33.28% probability that a randomly selected participant's response was between 4.5 and 6.5.

c) Find the probability that the mean of a sample of 16 selected participant's response was between 4.5 and 6.5.

Now we have $n = 16, s = \frac{2.3}{\sqrt{16}} = 0.575$

This probability is the pvalue of Z when X = 6.5 subtracted by the pvalue of Z when X = 4.5. So

X = 6.5

$Z = \frac{X - \mu}{\sigma}$

Due to the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{6.5 - 5.5}{0.575}$

$Z = 1.74$

$Z = 1.74$ has a pvalue of 0.9591.

X = 4.5

$Z = \frac{X - \mu}{s}$

$Z = \frac{4.5 - 5.5}{0.575}$

$Z = -1.74$

$Z = -1.74$ has a pvalue of 0.0409

0.9591 - 0.0409 = 0.9182

91.82% probability that the mean of a sample of 16 selected participant's response was between 4.5 and 6.5.