We are given that the lifetime of a battery in a certain application is normally distributed with mean μ = 16 hours and standard deviation σ = 2 hours.

Let X = lifetime of a battery in a certain application

So, X ~ N( $\mu=16,\sigma^{2} =2^{2}$ )

The z-score probability distribution for normal distribution is given by;

Z = $\frac{ X -\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean lifetime = 16 hours

$\sigma$ = standard deviation = 2 hours

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, the probability that a battery will last more than 19 hours is given by = P(X > 19 hours)

P(X > 19) = P( $\frac{ X -\mu}{\sigma}$ > $\frac{19-16}{2}$ ) = P(Z > 1.50) = 1 - P(Z $\leq$ 1.50)

= 1 - 0.9332 = 0.0668

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.50 in the z table which has an area of 0.9332.

Hence, the probability that a battery will last more than 19 hours is 0.0668.

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3 years ago

Hi please help me with this please help me draw the graph pls

sukhopar [10]

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