Respuesta :
Answer:
The percentage of lightbulb replacement requests numbering between 46 and 55 is 49.87%.
Step-by-step explanation:
Let the random variable X be defined as the number of daily requests to replace fluorescent lightbulbs.
The random variable X follows a Normal distribution with mean, μ = 55 and standard deviation, σ = 3.
To compute the probabilities of a Normally distributed random variable, we first need to convert the raw scores (i.e. X) to z-scores.
The formula to compute z-scores is:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
Compute the probability that the lightbulb replacement requests numbering between 46 and 55 as follows:
[tex]P(46<X<55)=P(\frac{46-55}{3}<\frac{X-\mu}{\sigma}<\frac{55-55}{3})[/tex]
[tex]=P(-3<Z<0)\\\\=P(Z<0)-P(Z<-3)\\\\=0.50-0.0013\\\\=0.4987[/tex]
*Use a standard normal table.
The percentage is, 0.4987 × 100 = 49.87%.
Thus, the percentage of lightbulb replacement requests numbering between 46 and 55 is 49.87%.
