the following histogram displays the distribution of battery life (in hours) for a certain battery model…

the following histogram displays the distribution of battery life (in hours) for a certain battery model used in cell phones: select one answer. 10 points suppose that battery life is a normal random variable with μ = 8 and σ = 1.2. using the standard deviation rule, what is the probability that a randomly chosen battery will last between 6.8 and 9.2 hours? a. 0.50 b. 0.68 c. 0.95 d. 0.997

the following histogram displays the distribution of battery life (in hours) for a certain battery model used in cell phones: select one answer. 10 points suppose that battery life is a normal random variable with μ = 8 and σ = 1.2. using the standard deviation rule, what is the probability that a randomly chosen battery will last between 6.8 and 9.2 hours? a. 0.50 b. 0.68 c. 0.95 d. 0.997

Answer

Explanation:

Step1: Recall the standard - deviation rule for normal distribution

The standard - deviation rule (also known as the 68 - 95 - 99.7 rule) states that for a normal distribution:

  • Approximately 68% of the data lies within 1 standard deviation of the mean, i.e., (P(\mu-\sigma<X<\mu + \sigma)\approx0.68).
  • Approximately 95% of the data lies within 2 standard deviations of the mean, i.e., (P(\mu - 2\sigma<X<\mu+2\sigma)\approx0.95).
  • Approximately 99.7% of the data lies within 3 standard deviations of the mean, i.e., (P(\mu - 3\sigma<X<\mu+3\sigma)\approx0.997).

Step2: Calculate the values of (\mu-\sigma) and (\mu+\sigma)

Given (\mu = 8) and (\sigma=1.2). (\mu-\sigma=8 - 1.2=6.8) and (\mu+\sigma=8 + 1.2=9.2). The probability (P(6.8<X<9.2)=P(\mu-\sigma<X<\mu+\sigma)).

Answer:

B. 0.68