Question
Suppose that variable X in a population has a distribution with mean 25 and standard deviation 4. If we choose a simple random sample of 50 people, what is the probability that the sample mean lies between 23 and 26? Show your approach.
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Answer to a math question Suppose that variable X in a population has a distribution with mean 25 and standard deviation 4. If we choose a simple random sample of 50 people, what is the probability that the sample mean lies between 23 and 26? Show your approach.
Timmothy
4.886 Answers
1. Calculez l'erreur standard :
\sigma_{\bar{X}} = \frac{4}{\sqrt{50}} \approx 0.5657
2. Trouvez les valeurs Z associées à 23 et 26.
Pour \( x = 23 \) :
Z_1 = \frac{23 - 25}{0.5657} \approx -3.54
Pour \( x = 26 \) :
Z_2 = \frac{26 - 25}{0.5657} \approx 1.77
3. Utilisez les tables de distribution normale pour trouver les probabilités :
P(Z < -3.54) \approx 0.0002
P(Z < 1.77) \approx 0.9616
La probabilité cherchée est :
P(Z_1 < Z < Z_2) = P(Z < 1.77) - P(Z < -3.54) \approx 0.9616 - 0.0002 = 0.9614
4. Conclusion : La probabilité que la moyenne de l'échantillon se situe entre 23 et 26 est approximativement 0.9614, or 96.14%.
2. Trouvez les valeurs Z associées à 23 et 26.
Pour \( x = 23 \) :
Pour \( x = 26 \) :
3. Utilisez les tables de distribution normale pour trouver les probabilités :
La probabilité cherchée est :
4. Conclusion : La probabilité que la moyenne de l'échantillon se situe entre 23 et 26 est approximativement 0.9614, or 96.14%.
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