Answer :
the probability that the sample will have a mean between 195 lbs and 200 lbs is approximately 0.7369 (rounded to four decimal places).
To find the probability that the sample mean falls between 195 lbs and 200 lbs, we'll first convert these values to z-scores using the formula:
[tex]\[Z = \frac{(X - \mu)}{\frac{\sigma}{\sqrt{n}}}\][/tex]
Where:
- [tex]\(X\)[/tex] is the value of interest (in this case, the sample mean),
- [tex]\(\mu\)[/tex] is the population mean,
- [tex]\(\sigma\)[/tex] is the population standard deviation, and
- [tex]\(n\) i[/tex]s the sample size.
For [tex]\(X = 195\, \text{lbs}\):[/tex]
[tex]\[Z_1 = \frac{(195 - 198.26)}{\frac{15.81}{\sqrt{147}}}\][/tex]
For [tex]\(X = 200\, \text{lbs}\):[/tex]
[tex]\[Z_2 = \frac{(200 - 198.26)}{\frac{15.81}{\sqrt{147}}}\][/tex]
Then, we'll look up the cumulative probabilities corresponding to these z-scores in the standard normal distribution table and calculate the probability that the sample mean falls between these two z-scores.
Let's calculate:
For [tex]\(X = 195\, \text{lbs}\):[/tex]
[tex]\[Z_1 = \frac{(195 - 198.26)}{\frac{15.81}{\sqrt{147}}} \approx -1.2689\][/tex]
For [tex]\(X = 200\, \text{lbs}\):[/tex]
[tex]\[Z_2 = \frac{(200 - 198.26)}{\frac{15.81}{\sqrt{147}}} \approx 0.9947\][/tex]
Now, we look up these z-scores in the standard normal distribution table to find the corresponding cumulative probabilities.
Using the standard normal distribution table:
- [tex]\(P(Z < -1.2689) \approx 0.1020\)[/tex]
- [tex]\(P(Z < 0.9947) \approx 0.8389\)[/tex]
Now, to find the probability that the sample mean falls between these two values, we subtract the cumulative probability of the lower z-score from the cumulative probability of the higher z-score:
[tex]\[P(-1.2689 < Z < 0.9947) = P(Z < 0.9947) - P(Z < -1.2689)\][/tex]
[tex]\[= 0.8389 - 0.1020\][/tex]
[tex]\[= 0.7369\][/tex]
So, the probability that the sample will have a mean between 195 lbs and 200 lbs is approximately 0.7369 (rounded to four decimal places).
