Respuesta :
Answer:
[tex]P (y < 6) = \frac{P(y- \mu)}{\sigma} <\frac{6> \mu}{\sigma} \\\\=P(z<\frac{6-6.88}{0.57} )\\\\=P(z<-1.54)\\\\=0.0618[/tex]
Therefore the probability that a randomly selected student has time for mile run is less than 6 minute is 0.0618
Step-by-step explanation:
Normal with mean 6.88 minutes and
a standard deviation of 0.57 minutes.
Choose a student at random from this group and call his time for the mile Y. Find P(Y<6)
[tex]\mu = 6.88[/tex]
[tex]\sigma = 0.57[/tex]
y ≈ normal (μ, σ)
The z score is the value decreased by the mean divided by the standard deviation
[tex]P (y < 6) = \frac{P(y- \mu)}{\sigma} <\frac{6> \mu}{\sigma} \\\\=P(z<\frac{6-6.88}{0.57} )\\\\=P(z<-1.54)\\\\=0.0618[/tex]
Therefore the probability that a randomly selected student has time for mile run is less than 6 minute is 0.0618
The probability that a randomly selected students has time for mile run is less than 6 minute is 0.0618.
[tex]P(y<6)=0.0618[/tex]
Given:
Number of able-bodied male students = 12,000
Normal with mean 6.88 minutes.
Standard deviation of 0.57 minutes.
μ = 6.88
σ = 0.57
y ≈ normal (μ, σ)
The z score is the value decreased by the mean divided by the standard deviation
[tex]P(y<6)=\frac{P(y-u)}{\sigma}<\frac{y>u}{\sigma}\\=P(z<\frac{6-6.88}{0.57}\\=P(z<-1.54)\\= 0.0618[/tex]
Therefore, the probability that a randomly selected students has time for mile run is less than 6 minute is 0.0618.
For more information:
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